sign in sign up
<<
Home , Limit of a sequence

Chapter 5

Limits for

Introduction Up until this point, we have mostly used limits without defining exactly what they are1. This approach matches the historical development. Indeed, scientist working in the 18th century, and earlier, such as , Oresme, Men- goli, Euler and even Newton, worked with limits without ever giving a proper definition for what this meant. In this chapter we follow the historical devel- opment into the 19th century when we define pre- Fig.1. The definition of the was cisely what we mean by the limit of a , actually first widely introduced into and use it to develop a proper theory that puts mainstream to more ef- our intuition on a firm mathematical ground. ficiently teach French soldiers calcu- 1Well, this is not really true, in Chapter 4.2, we tried lus in the early 1800s. to trick you into taking the definition of the limit out for aspin–seeexercises4.27,4.32,4.36and4.42

Remark 5.1 (Selected problems from previous exams based on this chapter)

1. Give the definition for limk ak =0, and use this definition to show that !1 sin k lim =0. k p !1 k

2. Give the definition for limk ak = , and use this definition to show that !1 1 n! lim = . k 20n 1 !1

209 210 CHAPTER 5. LIMITS FOR SEQUENCES

5.1 The definition for the limit of a sequence

Step 1 of 2: What it means for positive sequences to converge to zero

We begin by formulating the following special case of the definition of the limit in order to capture the main idea of what is going on in this chapter, and perhaps the main idea of what is going on in these lecture notes.

Definition 5.2 The positive sequence (an)n1=1 is said to converge to 0 if: For every ✏>0,thereexistsN>0 so that for all n N it holds that 2

n>N = an <✏. If this holds, we write )

lim an =0 or an 0 as n . n !1 ! !1 If (a ) does not converge to 0, then we write a 0. n n1=1 n 6! The Czech mathematician was the first to properly state this rather abstract definition of what we ought to mean by a limit. However, living in the mathematical backwater of Prague, it took around 50 years for anyone to notice (which also gave others – such as the French – the time to discover it independently). When you read the example below, notice how Definition 5.2 can be thought of as a game of Fig. 2. Bernard Bolzano (1781 – "challenges and responses". 1848). Never afraid of a challenge.

Example 5.3 We prove that 1 lim =0. n n !1 2 To do this, we need to show that the conditions of Definition 5.2 are satisfied when n an =1/2 . That is, we must prove that given any ✏>0, we can find a number N,so that 1 n>N = <✏. ) 2n (The hard part is usually to understand that this is what the definition asks of us.) While this proof can be done in one line, let us take our time to figure out what is going on. We begin by taking a specific epsilon, say ✏ =1/5. Since

21 =2, 22 =4, 23 =8, 24 = 16, 5.1. THE DEFINITION FOR THE LIMIT OF A SEQUENCE 211 it seems that if n>2, then we have 1/2n <✏. But what if ✏ is something else, such as ✏ =1/10? Well, in this case, by the same computation, n>3 implies 1/2n <✏.

Fig.3. In the language of the definition, for ✏ =1/5,wechooseN =2, and for ✏ =1/10,wechooseN =3. Notice how we can think of being challenged with an ✏ and then having to respond by giving a suitable N.

The point is this: to be able to say that a 0, we need to show that for all "✏- n ! challenges" (in the above figures, the challenge is represented by the height of the green line), we need to be able to respond by saying how far to the right do we have to go before every an is smaller than ✏. That is, we have to figure out how large N has to be so that an <✏whenever n>N. But how to do this for every ✏>0? Let us make two observations: It is hopeless to work with specific ✏, like we did above (that was just for fun). • However, to simplify life, we like to pretend that we can. Indeed, we usually assume that ✏ is a fixed, but unknown, number given to us by some person we respect, but are a little afraid of (like, say, my grandmother). This is supposed to remind us that we can use ✏ in our formulas – as if it was a concrete number – but that we are not allowed to change ✏ or assuming anything about it (except that it is strictly positive), since that would make grandmother angry. We cannot expect a specific N to work for every ✏. Indeed, the typical situation is • that the smaller ✏ becomes, the larger N has to be. In other words, N is going to be a of ✏ (sometimes we even write N(✏) to emphasise this). So, we assume that we have our concrete, but unknown, challenge ✏.Ourgoalisthe same as for the concrete values we considered above: to respond with an N.Todothis, we need to understand how large n needs to be for the inequality 1/2n <✏to hold. That is, we need to solve this inequality for n. This is exactly the type of thing we did in Chapter 0: 1 1 <✏ 2n > 2n () ✏ 1 log 2n > log () ✏ 212 CHAPTER 5. LIMITS FOR SEQUENCES

n log 2 > log ✏ () log ✏ n> . () log 2

That is, given ✏, a suitable choice for N is log(✏)/ log(2). Since this works for every n ✏>0, we have proved that by Definition 5.2 it holds that limn (1/2 )=0. !1

After reading the above example, you may be excused for thinking that the formula we found for N (as a function of ✏) is rather ugly and uninformative. But nothing could be farther from the truth. Indeed, let us see what happens for ✏ =1/5 and ✏ =1/10. This gives us: log ✏ n N ✏ =1/5= n =2,32... =2 n>2 ) log 2 )

log ✏ n N ✏ =1/10 = n =3,32... =2 n>3. ) log 2 )

But this is exactly what we got when we did this "by hand"! Moreover, the formula will tell us exactly which N we are supposed to choose no matter the ✏. Sort of nice, no?

Exercise 5.4 Answer the following questions on the above example.

(a) Use the formula to find suitable N when ✏ =1/1000 and 1/10000. (b) Use a while-loop to write a program in Python that for any given ✏ produces the first n so that an <✏. Verify that it matches what you found in (a). (c) As we are challenged with smaller and smaller ✏, does N become bigger or smaller? Does this match the formula for ✏ from the above example? Can you explain this graphically? (d) We seem to be focused on small ✏. What is a valid choice for N for large ✏? Say, ✏ = 100? Try to justify your answer using both your Python program, the inequality obtained above, and by thinking graphically.

Exercise 5.5 (a) Use the definition of the limit (as in Example 5.3) to prove 5 lim =0. n !1 2n +3 (b) Make a program using a while-loop in Python to verify that your formula for N in part (a) makes sense.

Exercise 5.6 Here is an exercise that at first may seem awkward. Prove that 1 lim =0. n !1 n! 5.1. THE DEFINITION FOR THE LIMIT OF A SEQUENCE 213

In the next example, now consider a sequence that has no limit.

Example 5.7 We show that the sequence a =1+( 1)n does not converge to zero. n This formula may look a bit strange, but if we check the first few numbers, we notice that (an)n1=1 =(0, 2, 0, 2, 0, 2, 0,...). Intuitively, it should be quite clear that the sequence does not converge to zero. But how to prove this? Well, suppose that we are challenged with ✏ =1. Then the point is that no matter how large we choose N,we can always find a number n>Nso that an >✏. For instance, this holds if we choose n =2N. Therefore, we have found one value of ✏ for which the definition cannot hold, and so it is impossible that an 0. (Notice, we ! Fig. 4. Here, ✏ =1. How far do we have have not proven that this sequence diverges to move to the right so that the a <✏? – which it does – but only that it does not n converge to 0.)

Exercise 5.8 We are given two sequences an (red) and bn (blue) as in Figure 5.

(a) For what value of N does it seem that n>N = a + b <✏for the indicated ) n n epsilons?

(b) Suppose that both an and bn converge to 0. Also, suppose that we are challenged with some unknown, but fixed, ✏>0.Explain why there must exist some number N so that an + bn <✏whenever n>N.Try to be as concrete as possible so that your explanation counts as a proof.

Exercise 5.9 Here is a warm up for the next section:

(a) Give a definition for limn an =0when !1 all an are negative.

(b) Give a definition for limn an =0when Fig. 5. !1 the an may be both positive and negative. 214 CHAPTER 5. LIMITS FOR SEQUENCES

Step 2 of 2: The general definition of the limit for sequences

We now explain how to pass from the spe- cial case of the definition of limits to the general case. In the general definition of the limit, what we really want to capture mathematically is what we really mean by

a L as n . n ! !1

But this should be the same as saying that a L approaches the value 0 as n in- n creases. Since we do not care about the sign of a L, we introduce absolute val- n ues, and this becomes

a L 0 as n . | n |! !1

But now a nice happens: if you apply the Fig. 6. Top: Here, some sequence an that definition of what it means for the positive appears to be approaching the limit L as n a L 0 sequence n to approach , you get grows. Bottom: That a L approaches | | | n | exactly the general definition for the limit. zero means that the length of the green lines That is, we get the following definition: on the figure to the left go to zero as n grows.

Definition 5.10 (The limit of a sequence) The sequence (an)n1=1 is said to converge to the finite value L if the following holds

For every ✏>0,thereexistsN>0 so that for all n N it holds that 2 n>N = a L <✏. )|n |

If this holds, we write

lim an = L or an L as n . n !1 ! !1

If an does not converge to any finite value L, then we say that the sequence diverges.

Just like like in the special case (Definition 5.2), the general definition tells us that to prove that a L, we must be able to win a game of challenges and responses. Indeed, n ! for every ✏>0 we are challenged with, we need to be able to respond with a number 5.1. THE DEFINITION FOR THE LIMIT OF A SEQUENCE 215

N (that depends on ✏) telling us how far we have to go to the right before the distances a L become less than ✏. | n |

Fig. 7. Left: The sequence from Figure 6 is challenged by an epsilon. Right: Choos- ing N =6, it seems we successfully responded to this challenge.

As we noted in Example 5.3, please keep in mind that N is a function of ✏. When we get challenged with smaller and smaller ✏, we usually have to respond with larger and larger N to compensate. In a sense, it really is a game of cat and mouse. Let us consider an example. Notice that the only difference from what we did in Example 5.3 is that we need to involve absolute values.

Example 5.11 We prove that n lim =1. n !1 n +1 According to the definition above, we need to show no matter which ✏>0 we are challenged by, then by choosing n large enough, we get n 1 <✏. n +1 Notice that the situation is exactly the same as in Example 5.3, except that the sequence 1/2n has Fig. 8. The sequence an = n/(n + 1). been replaced by the sequence (n/n + 1) 1 . | | So, what do we do? Well, the first thing is to try to simplify the expression for the terms: n n n +1 1 1 1 = = = . n +1 n +1 n +1 n +1 n +1 This means that the question has become: given ✏ >0, fixed but unknown, find a formula for how large n must be for us to have 1/(n + 1) <✏. Well, this is something we (should) 216 CHAPTER 5. LIMITS FOR SEQUENCES know how to do: 1 1 1 <✏ n +1> n> 1. n +1 () ✏ () ✏ Again, we have found a formula for how to choose n, and again, we are done!

Exercise 5.12 Prove that the sequence of partial sums Sn from the Geometric k k1=0 1/2 converges to 2. Hint: Why not try to get rid of the absolute value in the expression S L ? P | n | Let us now consider a slightly more complicated example.

Example 5.13 We prove that

2n2 3n +5 lim =2 n n2 9 !1 According to the definition above, we need to show no matter which ✏>0 we are challenged by, then by choosing n large enough, we get

2n2 3n +5 2 <✏. n2 9 Fig. 9. The sequence an = So, what do we do? Well, we compute! The first (2n2 3n + 5)/(n2 9). Notice that thing is to simplify the expression inside of the something weird happens at n =3. absolute value: Do you see why this is?

2n2 3n +5 2n2 3n +5 2n2 + 18 3n + 23 2 = = | |. n2 9 n2 9 n2 9 | | Keep in mind that our goal is to convince ourselves that forn large enough, this expres- sion is smaller than ✏. In particular, if n 8, then 3n + 23 =3n 23. Moreover, | | for n 78,italsoholdsthat n2 9 = n2 9. This allows us to simplify the fraction | | as follows: 3n + 23 3n 23 | | = . n2 9 (n 3)(n + 3) | | The next natural step would be to try to solve the inequality we obtain by putting this less than ✏. However, this would lead to rather annoying computations. To avoid this, let us try to simplify this expression further. We can do this as follows: for the denominator, we observe that 3n 23 3n. Moreover, for the numerator, we observe that n +3 n  (which gives 1/(n + 3) 1/n). In combination, this gives  3n 23 3n 3 = . (n 3)(n + 3)  (n 3)n n 3 5.1. THE DEFINITION FOR THE LIMIT OF A SEQUENCE 217

Finally, this expression is friendly enough for us to solve: 3 3 <✏ +3

n 8 2n2 3n +5 3 3 = 2 <✏. 8n> +3 ) n2 9  n 3 < ✏ In particular, this: means that whatever ✏>0 we are challenged with, we beat this epsilon-challenge by choosing n larger than the maximum of 8 and 3/✏ +3. We can formulate this by saying that given ✏>0,wechooseN = max 8, 3/✏ +3 . { }

Exercise 5.14 (a) In the previous example, we see in the figure that something strange happens at n =3. Why is this, and is this a problem for our computations and/or our final answer? (b) Write a program in Python that checks whether or not the formula for N in the above example makes sense. That is, the program should, given a value of epsilon, compute N according to the formula in the example, and then check how close the sequence is to the value 2 for a couple of n larger than N. Do you need to take the observation from (a) into account? Remark: For this exercise, the Python function np.ceil() can be useful. Given afloatingpointnumber,thisfunctionreturnsthesmallestintegerlargerthanit.

Exercise 5.15 (Discussion) Is it possible to write a Python program to prove that some limit holds? What about writing a program to show that a limit does not hold? Exercise 5.16 Prove that 6n n2 n +1 (a) lim =3 (b) lim =1 n 2n +5 n n2 + n 1 !1 !1 Hint: For (a), you can more or less do as in the first example above, but for (b), you need to follow the technique of the second example.

Exercise 5.17 Prove that 5n lim =0. n !1 n! Hint: Here, the key is to write out what the denominator and numerators really mean. 218 CHAPTER 5. LIMITS FOR SEQUENCES

5.2 The rulebook for limits of sequences

You may be happy to hear that when com- puting limits in practice, we rarely need to work the actual definition. Instead, we usually rely on softer arguments that we can justify thanks to a "rulebook" for lim- its that we establish in this chapter. In fact, the real reason for why we have been using the definition to check limits above, was merely that we wanted to take our new piece of mathematical technology out for a Fig.10. Only after taking your new toy spin. out for a test-drive would you trust it to take you where no man has gone before! To get a feeling for what the rulebook for limits will look like, we start by doing some intuitive computations below. We will see a situation where our intuition works, and one where it fails.

An example where a naive intuitive approach works In the following example, we illustrate the typical intuitive approach used by mathe- maticians when computing limits. The main idea of this section is to both justify this approach and explain its limitations.

Example 5.18 In exercise 5.16 we used the definition of the limit to check that 6n lim =3. n !1 2n +5 However, to actually figure out that the limit is equal to 3, you may agree that the following intuitive computation seems quite convincing:

6n n large 6n lim lim =lim3=3. n n n !1 2n +5 ⇡ !1 2n !1 Here, we simply say that when n is of the size, let us say a few gazillions (to use a technical term from Donald Duck), then the +5 in the denominator can safely be ignored. To support this conclusion, we can use Python either to plot a few thousand terms, or to simply compute what happen for a few large values of n by using the command print(6 n (2 n+5) ( 1)). ⇤ ⇤ ⇤ ⇤⇤

Exercise 5.19 The same approach used in the above example should work to deter- mine the following limits. (n + 10)(n + 100) p9n6 +1+n3 n2 +sin(3n + 1) (a)lim (b)lim (c)lim n 3 n 3 n 2 n !1 100(n + 1) !1 2n +5 !1 3n + cos(3 + 2) 5.2. THE RULEBOOK FOR LIMITS OF SEQUENCES 219

An example where the same naive intuitive approach fails

While the method of example 5.18 is convenient, it is easy to slip up when you do not actually know what you are doing2.

Example 5.20 We consider the limit n2 lim n . n n +2 !1 ✓ ◆ Arguing as in Example 5.18, it is intuitively clear that for large n, we can ignore the +2 in the denominator. This implies that n2 n2 = n, n +2 ⇡ n which again leads to n2 lim n lim (n n)= lim 0=0. n n +2 ⇡ n n !1 ✓ ◆ !1 !1 Case closed! No? Ok, just to be sure, let us use Python to visualise what is going on:

Fig. 11. To the left we have the first 20 terms. They do not seem to go to zero. To the right, we include the first 100 terms. Still no sign of tending to zero.

Whops! It seems that 2,andnot0, is the correct limit for the sequence. Let us see what happens if we rewrite the expression before computing the limit: n2 n(n + 2) n2 2n n = = . n +2 n +2 n +2 We now use the same intuition as above, namely that n +2 n for large n. This gives: ⇡ n2 2n 2n lim n =lim lim =2. n n +2 n n +2 ⇡ n n !1 ✓ ◆ !1 !1 Well, that is convenient. This time the answer matches Python’s suggestion.

2Which is why, at the end of the day, we need a formal definition for the limit to separate between what is correct and what is not. 220 CHAPTER 5. LIMITS FOR SEQUENCES

Arulebookforcomputinglimits

Our goal now is to figure out what are allowed to do when computing limits. Our plan is as follows. First, we state the entire "rulebook" for computing limits. Second, we play around with these rules in order to understand how to use them. And finally, we prove them.

Proposition 5.21 (Rulebook for the limit of sequences) Suppose that the limits limn an and limn bn exist (and are finite). Then the following hold: !1 !1

(i) lim an bn =liman lim bn n ± n ± n !1 ⇣ ⌘ !1 !1 (ii) lim an bn =liman lim bn n n !1 · · !1 If, in addition, limn bn =0, then !1 6

(iii) lim an/bn =liman / lim bn n n n !1 !1 !1 The following rule is rather famous, and is often called the :

(iv) If a b c ,andlim a =limc = L,thenlim b = L. n  n  n n n n The following rule cannot really be understood until we reach Chapter 6:

If f is continuous and both the an and its limit (v) are contained in Df ,then lim f(an)=f(lim an) ( n n !1 !1 Moreover, the following concrete limits are useful enough to be included here: 1 (vi) lim C = C (vii) lim =0. n n !1 !1 n Here C is a constant.

Exercise 5.22 (a) Prove rule (vi),and(b) prove rule (vii), above. Exercise 5.23 Prove that a 0, b 0= a + b 0. n ! n ! ) n n ! Remark: This just asks you to write the solution of exercise 5.8(b) properly. The reason why rule (v) cannot really understood right now is that we do not yet know what it means for a function f to be continuous. We also mention that some textbooks state this rules in a slightly different way which avoids mentioning continuity:

(v’) lim an = A and lim f(x)=B = lim f(an)=B. n x A ) n !1 ! !1 (We remark that to make this rule look slightly nicer than it actually is, we are skipping one little technicality. We point it out in the next chapter.) 5.3. HOW TO USE THE RULEBOOK IN PRACTICE 221

5.3 How to use the rulebook in practice

Indeterminate forms

The first step to understanding how to use the rulebook for limits is to notice that every rule requires that all the limits involved exist and are finite numbers. For this reason, it is important to be able to know when this is not the case. To keep track of this, we introduce a notation for so-called indeterminate forms. To explain what this is, we recall the following two expression that we considered in Example 5.20: n2 2n lim n and lim . n n +2 n n +2 !1 ⇣ ⌘ !1 If we consider the second expression first, then notice that as n grows, then both the numerator and denominator become infinitely large. We say that this expression is of the type [ / ] and write 1 1 2n lim = 1 . n n +2 !1 1 Here, we use square brackets around the " / h" toi remind us that is not an actual 1 1 1 number (if it was, we could write / =1and be done – but notice that in the example, 1 1 this quotient should be equal to 2). In the first expression above, the main "struggle" is between the term n and the quotient n2/(n + 1), since, as n grows, both become "infinitely" large. We express this by writing n2 lim n =[ ]. n !1 n +2 11 Again, we use square brackets to⇣ remind us that⌘ is not an actual number, and there 1 is no reason to expect that =0(this is the mistake we made in the example!). 11

Remark 5.24 (indeterminate forms) Limit expression to which the rulebook (or any reasonable extension of it) does not immediately apply are called indeterminate forms. In addition to what we saw above, these can be of the form [0/0], [0 ], [ 0], [1 ] or ·1 1 1 even [00].

Exercise 5.25 Here is an example of a limit of the [11]: 1 n lim 1+ . n n !1 ⇣ ⌘ Use Python to figure out what the limit of this expression is (we will revisit it below.) Exercise 5.26 (discussion) (a) Define what we mean when we say that a sequence tends to infinity. That is, define what we mean by a as n . (b) Suppose n ! 1 !1 that a and b . Use your definition to prove that then also a + b . n !1 n !1 n n !1 Remark: You have now proved that [ + ] is not an indeterminate form (since it is 1 1 always equal to ). 1 222 CHAPTER 5. LIMITS FOR SEQUENCES

Rule of thumb 1: Rewrite your expression as a quotient

In practice, most limits we consider are indeterminate forms of some type. So how to deal with them? Well, if you look back at Example 5.20, notice how we rewrote an expression on the form [ ] so that it become on 11 the form [ / ]. As a rule of thumb, rewriting your ex- 1 1 pression on the form of a quotient usually makes it easier to solve. Let us revisit Example 5.20 to illustrate how this rule Fig.12. Quotients good! of thumb helps us out.

Example 5.27 We now break the computation of the limit from Example 5.20 into steps that can be justified. First, we notice that

n2 lim n =[ ]. n n +2 11 !1 ⇣ ⌘ Following the rule of thumb, we begin by rewriting this expression on the form [ / ]: 1 1 n2 n(n + 2) n2 lim n =lim n n +2 n n +2 !1 ⇣ ⌘ !1 2n =lim = 1 . n n +2 !1 h1i So, what next? Notice that by rules (vi) and (vii) in the rulebook, we know the limits of constants and the expression 1/n. To get these into play, let us multiply by one: 2n 2n (1/n) 2 lim =lim =lim . (5.2) n n n 2 !1 n +2 !1 n +2· (1/n) !1 1+ n Finally, we have arrived at an expression that we can apply our rulebook to: lim 2 2 n lim = !1 n 1+ 2 lim (1 + 2 ) !1 n n n !1 (5.3) 2 2 = = =2. 1+ lim 2 1+0 n n !1

Notice the following, rather peculiar, detail in the above example. It is only after making the very last step of the computation (5.3), that we know that the preceding steps in this computation are ok. Only then do we know that all limits involved exist. 5.3. HOW TO USE THE RULEBOOK IN PRACTICE 223

Rule of thumb 2: In a quotient, let the strongest terms fight eachother

In the above example, we multiplied by one to get into a situation where we could use the rules from the rulebook. Another way of thinking about this, which is more flexible, and therefore more useful, is in terms of factoring out the strongest terms. That is, figure out who is the strongest in the numera- tor and denominator, respectively, and factor these terms out. Chances are that this will leave you with a computable expression. Fig.13. Think of this as choosing Here is an example of how this works. champions who are to fight.

Example 5.28 Let us compute the limit n2 +1 lim . n 3 !1 3n +2n +1 This limit is already on the form [ / ]. Here, we see that the strongest term in 1 1 the numerator is n2, while the strongest term in the denominator is n3 (we ignore the constant). By factoring out these terms, we arrive at the following computation:

2 2 1 1 n +1 n 1+ 2 1 1+ 2 1+0 lim =lim n =lim n =0 =0. n 3n3 +2n +1 n n3 2 1 n n 2 1 3+0+0 !1 !1 · 3+ n2 + n3 !1 · 3+ n2 + n3 · In the last step, we merely indicate how to use the rulebook to finish the computation.

Notice that if we flipped things around, we would get the computation 3 2 1 3n +2n +1 3+ 2 + 3 lim =limn n n =[ 3]. n n2 +1 n 1 !1 !1 · 1+ n2 1· The problem is that the rulebook does not actually tell us how to deal with this situation. However, it ought to be intuitively clear that it is safe to say that the above expression is not really indeterminate since [3 ]= . The following exercise indicates several ·1 1 formulas that we can add to our rulebook:

Exercise 5.29 Combine the two first rules of thumb to quickly verify all the limits in exercise 5.19. Exercise 5.30 Interpret the following formulas in a suitable way, and prove them: (a) [C + ]= (b) [C ]= 1 1 ·1 1 Remark: This is a follow-up to exercise 5.26. Moreover, in (b), you need to add some conditions on what C is allowed to be. 224 CHAPTER 5. LIMITS FOR SEQUENCES

Rule of thumb 3: The limit loves smooth functions

This rule of thumb is just our way of trying to make sense of computational rule (v), which we do not really have the background to understand yet. Specifically, the rule says that the formula

lim f(an)=f lim an n n !1 ⇣ !1 ⌘ holds if the function f is continuous. Being con- tinuous essentially means that the graph of the function does not have any "abrupt" jumps on Fig. 14. Hey! Is this allowed? As we its domain. In particular, the graphs of all func- will see in Chapter 6, this rule is even tions we saw in Chapter 2 are nice and smooth on more forgiving than what we formu- their domains and so satisfy this rule of thumb. late here.

Example 5.31 Since the graph of the function is nice and smooth, we can use the rule of thumbs as follows: n2 n2 lim n = lim n = p2, n n +2 n n +2 !1 r⇣ ⌘ r !1 ⇣ ⌘ where, in the last step, we used the result of Example 5.27.

Exercise 5.32 Compute the exact value of both of the following limits: 1 (a) lim cos sin (b) lim cos(arctan n) n n n !1 ⇣ ⌘ !1 Exercise 5.33 Use the trick a2 b2 =(a + b)(a b) to compute the following limits. 1 (a)lim (b)lim n2 + n n n p 2 p 2 n !1 n + n n +2 !1 ⇣p ⌘ Exercise 5.34 Does the following computation work? Compare with exercise 5.25. 1 n 1 lim 1+ =limexp n ln 1+ n n !1 n !1 · n ⇣ ⌘ ⇣ 1⌘ =exp lim n ln 1+ =exp lim n ln 1+0 =exp lim 0 =1. n n n !1 · n !1 · !1 ⇣ ⌘ ⇣ =0 ⌘ ⇣ ⌘ Exercise 5.35 (Challenge) Use Definition 5.10 to prove a| {zA =} pa pA. n ! ) n ! 5.3. HOW TO USE THE RULEBOOK IN PRACTICE 225

Rule of thumb 4: When all else fails, give your limit a good squeeze This rule of thumb is our way of emphasising that rule (iv), called the Squeeze theorem, turns out to be extremely useful. Another indication of its usefulness is that it has many nicknames. For instance, it has been called the "Sandwich the- orem", the "Policeman theorem" and the "KGB theorem" (apparently the last one was popular among soviet mathematicians). In Sweden, it is usually called "Instängningssatsen". The squeeze theorem is a rather important rule, since it allows us to compute limits that we, for some reason, find difficult by replacing Fig. 15. Or... call social services? them with a pair of simpler limits. Here is an example where the Squeeze theorem helps us out.

Example 5.36 We want to compute the limit sin n lim . n !1 n Since sin(n) never settles as n increases, this limit is not like any of the indeterminate forms we have seen before. To understand what is going on, we try to use the Squeeze theorem. Indeed, taking the in- equality 1 sin n 1,   and dividing it by n yields 1 sin n 1 . n  n  n Since both expressions on the "outside" tend to 0 as n , it follows, by the Squeeze theorem, !1 that the expression in the middle is "escorted" Fig. 16. Our original sequence is in to the same limit (see figure to the right). That blue. The red dots represent the up- is, by rule rule (iv), we conclude that per and lower bounds. sin n lim =0. n !1 n 226 CHAPTER 5. LIMITS FOR SEQUENCES

In Chapter 3, we mentioned that computers are not always to be trusted. For in- stance, below, we use Python to help us visually determine the limit. Let us now consider the limit n n lim 2 log(1 + 2 ). (5.4) n !1 Here is how Python visualises the behaviour of the sequence in question:

Fig.17. To the left, we see the first 20 terms, and visually, it appears that the limit is 1. However, when we visualise the first 80 terms on the right, we see a sudden drop to 0.

In the following exercises you are to study the limit (5.4), above.

Exercise 5.37 (a) The sudden drop to 0 in Figure 17 is due to a round-offerror. Explain why it seems to happen almost exactly at n = 53.(Hint: Recall exercise 3.71) To better understand the limit (5.4), we consider the inequalities

x 1 log x x 1. x>0. (5.5) x   8

(Recall Proposition 2.52.)

(b) Plot both curves y = log x and y = x 1 in the same coordinate system. Do the inequalities in (5.5) seem reasonable? n n (c) Apply the inequalities to the expression 2 log(1 + 2 ) in order to squeeze it between two more friendly sequences. What does this say about the limit?

Exercise 5.38 Use the inequalities from (5.5) to determine

1 n lim 1+ . n n !1 ⇣ ⌘ Hint: The connection to the is given in exercise 5.34. 5.3. HOW TO USE THE RULEBOOK IN PRACTICE 227

Aspecialnotationforthesqueezetheorem:TheBig-oh

As another indication of how useful the squeeze theorem is, we briefly mention the following notation designed to be used by this theorem:

Definition 5.39 (Big-oh notation) Let (an)n1=1, (bn)n1=1 be sequences. Then we say that a = (b ) if there exists a constant C>0 so that n O n

an C bn , n N. | | | | 8 2

Note that we can read the above definition as saying that the sequence an is dom- inated by the sequence bn (notice that we do not really care about the value of the constant C, since we are usually only interested in knowing if an is forced down to zero.). We will only really care about the Big-oh notation in the last chapter of these lecture notes (but then we will really care about the Big-oh). So for the moment, we only give the following exercise as a small taste of things to come:

Exercise 5.40 Suppose that you have a sequence (a ) where a = (1/n). n n1=1 n O (a) Explain why this means that an is squeezed between two inequalities. (b) What is the only reasonable value of the limit

lim an ? n !1 Exercise 5.41 Suppose that a = n2 . For which values of ↵>0 can we deter- n O mine the limit an lim ? n ↵ !1 n Exercise 5.42 Is it possible for one and the same sequence an to satisfy both, say, a = (n) and a = n2 at the same time? If so, give an example where this n O n O happens, if not, prove that it leads to an absurdity. Exercise 5.43 (a) Explain why n n2 k = + (n) . 2 O Xk=1 Remark: Here, you are allowed to use the result of Example 1.82. (b) Use the Big-oh notation to compute the following limits in as few steps as possible. n n n 1 1 1 (i) lim k (ii) lim k (iii) lim k n 3 n 2 n !1 n !1 n !1 n Xk=1 Xk=1 Xk=1 228 CHAPTER 5. LIMITS FOR SEQUENCES

Awordofadviceonhowtochooseyourchampion:thetableofgrowth

Rule of Thumb 2 is all about identifying the strongest term in the denominator and numerator in order to understand the indeterminate forms [ / ] and [0/0]. This is 1 1 important since most indeterminate forms we meet may be rewritten on this form.

Example 5.44 To rewrite [ 0], we essentially use that [1/0+]=[ ]. For example: 1· 1 1 1 log 1+ n 0 lim n log 1+ =[ 0] = lim = . n · n 1· n ⇣ 1 ⌘ 0 !1 ⇣ ⌘ !1 n h i

Notice that while you are not allowed to write 1/0= (division by zero is not 1 defined!), it is perfectly fine to write [1/0+]= . Indeed, by this we mean that if we 1 divide 1 by a sequence that goes to zero (from the positive side) this gives us something that goes to (positive) infinity. The following notation helps us identify the strongest term more efficiently.

Definition 5.45 We write an bn as n to indicate that limn an/bn =0. ⌧ !1 !1

The relative strengths of the most common sequences may now be stated as follows.

Proposition 5.46 (table of growth) For any 0 <↵<and 1

Fig. 18. Here you can compare the sequences from the table of growth. Note that n2 is represented in both plots as the red dots.

Exercise 5.47 Use the table of growth, in combination with the rules of thumbs above, to compute the following limit. en +2lnn +3n10 arctan(n)ln(n10)+4ln(n100) (a)lim (b)lim . n n 100 n 2 100 !1 5e +7lnn + 11n !1 cos(1/n )ln(n )+5ln(n) 5.3. HOW TO USE THE RULEBOOK IN PRACTICE 229

We now justify the table of growth in the following example and exercises.

Example 5.48 (Exponential functions versus ) We use the squeeze theorem to show that 3n n! as n . That is, we show that ⌧ !1 3n lim =0. n !1 n! n To apply the Squeeze theorem to the sequence bn =3/n!, we first notice that the expression is positive. This means that we always have the lower bound 0 3n/n! (and  that we can work without absolute values). All that remains is to find a suitable upper bound cn. To better see what is going on, we make the following computation: 3n 3 3 3 3 3 3 3 3 3 27 = 1 = . n! 1 · 2 · 3 · 4 ···n 1 ·n  1 · 2 · · n 2n 1  n We conclude by applying the Squeeze| {z theorem} with an =0,bn =3 /n! and cn = 27/2n.

Exercise 5.49 Show that the following is true as n . !1 (a) Given any fixed, but unknown, b>0,showthatbn n!. ⌧ (b) Show that n! nn. ⌧ (c) Use (a) and (b) to deduce we also have bn nn. ⌧ Hint: For (a) and (b), modify the above example.

Exercise 5.50 In this exercise, we prove that for all ↵>0,wehave

log n n↵ as n . (5.6) ⌧ ! 1 (a) Combine the squeeze theorem with a suitable inequality for the logarithm from exercise 5.37 to convince yourself that (5.6) holds for ↵ =2. For what other values of ↵ does the argument work? (b) Combine the inequality you used from exercise 5.37 with the identity log(n1/2)= (log n)/2 to show that (5.6) also holds for ↵ =1. (c) Modify your argument from (b) to show that (5.6) holds for all 0 <↵<1.

Exercise 5.51 Use the limit from the previous exercise to show that for all >0 and a>1,wehaven an as n . ⌧ ! 1 Hint: You need to rewrite the limit expression so that the logarithm appears. 230 CHAPTER 5. LIMITS FOR SEQUENCES

5.4 How to prove the rules in the rulebook

In this section we prove all the remaining rules in the rulebook, except rule (v).Forthe latter rule, we need to wait for the next chapter where we discuss limits for functions. Note that every proof ends with an exercise that you need to complete. This is to force you to try to actively think about the proofs and discuss them with others!

First things first: a proof that the limit is "well-defined"

Recall that Grandi was able to show that 1=0by using a "bad" definition for the sum of an infinite series. What if our definition of the limit has a similar weakness? Well, it does not. The following result says that our definition of the limit is "well-defined":

Proposition 5.52 If the sequence (an)n1=1 has both the limits L and M, then L = M.

Proof. To prove this by contradiction, we suppose that we have two numbers L = M 6 such that both a L and a M n ! n ! at the same time. According to Definition 5.10, this means that for every ✏1 > 0 and ✏2 > 0, there exist numbers N1 and N2 so that n>N = a L <✏ , and (5.7) 1 )|n | 1 n>N = a M <✏ . (5.8) 2 )|n | 2 That is, we can get the distance between an and L,andan and M,tobeassmallaswe want, provided that we make n large enough. Keep in mind, that the distance between L and M is fixed. Here is an illustration of this (absurd) situation:

Fig. 19. We can get the distances labeled I and II as small as we want, but the distance III is of fixed length. This makes no sense at all! In the exercise below, you are asked to verify this.

Exercise 5.53 Based on the intuition given by the above figure, try to deduce an absurdity by choosing ✏1 and ✏2 sufficiently small. Hint: Since you are comparing distances between three points, the x + y x + y may be useful. | || | | | 5.4. HOW TO PROVE THE RULES IN THE RULEBOOK 231

Proof of Proposition 5.21(i): The addition rule

We consider the proof of the summation rule for the limit. Notice that this proof is essentially a careful solution to exercise 5.8 carefully. We begin by recalling the rule we seek to prove:

Proposition 5.21(i) (slightly reformulated)

lim an = L and lim bn = M = lim (an + bn)=L + M. n n n !1 !1 ) !1

Exercise 5.54 The biggest challenge here may actually be to understand why there is something to prove here. To shed a little light on this, consider the sequences a =1/n +( 1)n and b =1/n ( 1)n. n n (a) Determine whether or not the following limits exist:

(i) lim an (ii) lim bn (iii) lim (an + bn). n n n !1 !1 !1 (b) What does this example tell you about computational rule (i) for limits?

Proof of Proposition 5.21(i) To prove that a + b L + M, we begin assuming we are n n ! given an ✏>0 that is fixed but unknown (this philosophy was explained in, for instance, Example 5.3). According to the definition of the limit, we must respond to this epsilon by showing that there exists some number N so that the following implication is true:

n>N = (a + b ) (L + M) <✏. (5.9) )| n n | To do this, we need to use the hypothesis of the computational rule we are trying to prove. According to the definition of the limit, the statements a L and b M n ! n ! mean the following: for all ✏1,✏2 > 0, there exist numbers N1,N2 > 0,sothat

n>N1 = an L <✏1 )| | (5.10) n>N = b M <✏ . 2 )|n | 2 (Here, we use subscripts to keep these epsilons apart from each other and from the epsilon we are given above.) Having lined up all these facts, we are almost done. We ask you to end the proof in the following exercise.

Exercise 5.55 (a) Use the triangle inequality to show that (a + b ) (L + M) | n n | a L + b M . | n | | n | (b) We know that (5.10) holds for all choices of ✏1,✏2. For which choice does this, in combination with (a), imply (5.9)? (Now is the time to use the given ✏!) (c) What is a suitable choice for N? (Here it may be of help to recall exercise 5.8.) 232 CHAPTER 5. LIMITS FOR SEQUENCES

Proof of Proposition 5.21(iv): The squeeze theorem

We now turn to the proof of the Squeeze theorem.

Proposition 5.21(v) If a b c , then the following holds: n  n  n

lim an = L and lim cn = L = lim bn = L. n n n !1 !1 ) !1

(Note that we do not need to assume that the limit of the sequence bn exists.)

Proof of the Squeeze theorem As usual, we begin by assuming that we are given a number ✏>0 which is fixed but unknown. Our goal is to respond to this epsilon by showing that there exists some number N so that the following implication is true:

n>N = b L <✏. (5.11) )|n | According to the hypothesis, what we know is that for all ✏1,✏2 > 0, there exist numbers N1,N2, respectively, such that

n>N1 = an L <✏1, )| | (5.12) n>N = c L <✏ . 2 )|n | 2 Moreover, by the hypothesis, we have the following connection between the three se- quences an,bn,cn. Namely, for all n it holds that

a b c . (5.13) n  n  n To take advantage of this connection, we use the result of exercise 2.48 to “open up” the inequalities in (5.12) and rewrite them as follows:

an L <✏1 L ✏1

Exercise 5.56 (a) Combine (5.13) and (5.14) to obtain a double inequality for b n L.

(b) We know that (5.12) holds for all choices of ✏1,✏2. For which choice does this, in combination with (a),imply(5.11)? (c) What is a suitable choice for N?

Exercise 5.57 Prove the following useful variation of the Squeeze theorem: If a b n  n for all n N,andan A and bn B as n . Then A B. 2 ! ! !1  5.4. HOW TO PROVE THE RULES IN THE RULEBOOK 233

Proof of Proposition 5.21(ii),(iii): The and quotient rules

We start out by considering the multiplication rule in a slightly reformulated form:

Proposition 5.21(ii) (slightly reformulated)

lim an = L and lim bn = M = lim (anbn)=LM. n n n !1 !1 ) !1 The proof begins exactly as in that of the two previous ones: Exercise 5.58 As in the previous two proofs, write out what the hypothesis and conclusion of Proposition 5.21(ii) means in terms of the definition of the limit. Next, we need to somehow use the information that a L and b M to conclude n ! n ! that a b LM. That is, we need to relate the distances a L and b M to the n n ! | n | | n | distance a b LM . (This step is analog to part (a) of exercises 5.55 and 5.56.) | n n | We do this by using the trick of adding by 0 followed by the triangle inequality: a b LM = a b a M + a M LM | n n | | n n n n | = a b a M + a M LM | n n n n | (5.15) = a b M + M a L | n n n | a b M + M a L. | n|| n | | || n | This seems rather promising! Indeed, it would seem that we could now end the proof by choosing ✏ = ✏/2 M and ✏ = ✏/2 a . However, life is not this good. By the definition 1 | | 2 | n| of the limit, the epsilons have to be constants, and in particular, are not allowed to depend on n. Therefore our choice of ✏2 is not valid. We invite you to fix this in the following exercise. (We also have a problem if M =0. Do you see how to fix this?) Exercise 5.59 (a) Show that there exists an N such that for all n>N then a 3 3 | n| L +1. (Hint: Try to understand this by making an appropriate illustration.) | | (b) Combine your inequality in (a) with the computation (5.15). Use the resulting inequality to choose suitable ✏ and ✏ so that you end up with a b LM <✏. 1 2 | n n | (c) What is a suitable choice for N? Exercise 5.60 (Challenging) After some preparation, the follows from the multiplication rule. In this exercise, we ask you to go through the details. (a) Prove the reverse triangle inequality: a + b a b . | || || | (b) Prove that b M implies that b M /2 for sufficiently large n. (This is n ! | n|| | analog to part (a) of the previous exercise.) (c) Prove that b M implies 1/b 1/M . (You need part (b) here.) n ! n ! (d) Use the multiplication rule for the limit to conclude. 234 CHAPTER 5. LIMITS FOR SEQUENCES

Proof of Proposition 4.44: the Balloon lemma

To prove part (i), suppose that cn is growing and bounded from above. That the sequence is grow- ing means exactly that cn cn+1 for all n N,  2 and that the sequence is bounded from above means exactly that there exists some number M so that cn M for all n N (see Figure 20).  2 Our first job is to find something for the se- quence to converge to. As should be clear from the figure, there is no reason to expect the se- quence to converge to M. However, from the supremum axiom, we know that since the set

C = c ,c ,c ,... { 1 2 3 } is bounded above, it also has a smallest upper bound sup M. We denote it by L. The idea is now to try to prove that c L. n ! As in the above proofs, we start out by letting ✏>0 be a fixed but unknown number. Opening up the absolute values, we can state that our goal c is to find a number N so that Fig. 20. Top: The sequence n with its upper bound M, and supremum n>N = L ✏L ✏. This is great, since the existence of this number is the crucial 0 n0 ingredient which allows us to wrap us this proof (as we ask you to do in the following exercise).

Exercise 5.61 (a) Explain why there exists an index n so that c >L ✏. 0 n0 (b) Explain why, for all sufficiently large n,wehavec >L ✏. n (c) Explain why, for all n,wehavecn N = c L <✏true? )|n | Exercise 5.62 Adapt the above argument into a proof of part (ii) of the proposition. 5.5. CONVERGENCE TESTS FOR POSITIVE SERIES 235

5.5 Three convergence tests for positive series using limits

We end this chapter with a section aimed at motivating, formulating and finally prov- ing three additional convergence tests for series that, in some sense, are slightly more advanced than the and comparison tests that we saw in Chapter 4.

The

To motivate the test, we consider the following example.

Example 5.63 (Part 1 of 2) Let us check whether the following series converges or diverges: 1 pk2 + k +1 . k2 +1 Xk=1 As we noted in Chapter 4, what is important is the behaviour of these terms for large values of k. Intuitively, we see that

pk2 + k +1 large k pk2 +0+0 1 = . (5.16) k2 +1 ⇡ k2 +0 k This means that for large k, the series behaves like the harmonic series (which is di- vergent). For this reason, we expect this series to diverge as well. However, to justify this using the usual comparison test (Proposition 4.50), we would now need to establish an inequality relating the terms of these series with each other. The point of the limit comparison test is that we can avoid the inequalities. We now put this example on hold to discuss how to do this.

Exercise 5.64 Complete the above example using the usual comparison test (Propo- sition 4.50).

Let us now take a moment to consider what limit could be relevant for the above example. The intuitive computation (5.16) is essentially a naive version of the following computation (which is justified, step by step, by the rulebook for limits):

pk2 + k +1 k2 +1 k(pk2 + k + 1) lim =lim = 1 k ⇣ 1 ⌘ k k2 +1 !1 !1 h1i k (5.17) ⇣ ⌘ 2 1 1 k 1+ k + k2 p1+0+0 =lim = =1. k k2 q 1+ 1 1+0 !1 k2 =1 |{z} 236 CHAPTER 5. LIMITS FOR SEQUENCES

In Figure 21, we visualise the above computa- tion. We see that the terms of the original series (blue) and the terms of the the harmonic series (red) are quite close to eachother. In green, we visualise the quotient between these terms. As we verified in the computation (5.17), these quo- tients tend to the value 1. In particular, this means that if we take k large enough, then the quotient between the terms of the original series and the harmonic se- ries (the green dots) will lie as close to 1 as we want. In particular, for ✏ =0.1, there exists an Fig. 21. N so that pk2 + k +1 2 k>N = 0.9 k +1 1.1 )  ⇣ 1 ⌘  k Multiplying up the factor (1/k), this becomes ⇣ ⌘

0.9 pk2 + k +1 1.1 k>N = . ) k  k2 +1  k But this gives us exactly what we need to apply the usual comparison test (Prop. 4.50). In this case, we want to show divergence, so by the left-most of these two inequalities, we conclude that our original series diverges. (Recall that a series converges if and only if its tail converges. This also means that a series diverges if and only if the same is true for its tail.) We have now, more or less, indicated how to prove the following result.

Proposition 5.65 (Limit comparison theorem for infinite series) Suppose that a ,b are two positive sequences, and that there exists L 0 so that k k a lim k = L. k bk If 0

We now return to Example 5.63. Notice how the intuitive and formal steps are now almost identical. 5.5. CONVERGENCE TESTS FOR POSITIVE SERIES 237

Example 5.63 (Part 2 of 2) We study whether the following series converges or not:

1 pk2 + k +1 . k2 +1 Xk=1 Intuitive step: Do as in Example 5.63, and observe that for large k, the terms in this series behave like those of the harmonic series k1=1 1/k. This means we expect divergence. P Formal step: We repeat, word by word, the computation (5.17) above. This shows that the limit comparison test with L =1applies. We conclude that our series is divergent since the same holds true for k1=1 1/k. P Exercise 5.66 Solve exercise 4.60 using the limit comparison theorem.

Exercise 5.67 Solve exercise 4.61 using the limit comparison theorem.

Exercise 5.68 Use the limit comparison test to determine if the following series con- verge or diverge.

1 n + log n 1 3n + n4 (a) (b) . n2 p n 3 n=1 n=1 4 + n X X Hint: It may be a good idea to use the table of growth here.

Exercise 5.69 Turn the discussion leading to Proposition 5.65 into a proper proof for the case that 0

Exercise 5.70 Explain intuitively what should happen when L =0and L = in 1 Proposition 5.65, respectively. Explain how to modify the proof in each of these cases.

Exercise 5.71 (Challenge) Determine for all ↵ R whether the series 2 1 1 ( n k)↵ n=1 k=1 X converges or diverges. P 238 CHAPTER 5. LIMITS FOR SEQUENCES

The

We explain the basic idea of the root test in the following example.

Example 5.72 Suppose that 1 a is a positive series where, for all k 1,wehave k=1 k the estimate P 1 a1/k . k  2 But this is equivalent to 1 a . k  2k k Since k1=1 1/2 is a convergent , it follows by the standard comparison test that the series 1 a is convergent. P k=1 k P The general formulation of the root test is as follows.

Proposition 5.73 (Root test) Suppose that ak is a sequence of positive numbers, and that

k lim pak = Q. k !1 Then the following hold:

(i) Q<1= 1 a converges ) k=1 k (ii) Q>1= P1 a diverges ) k=1 k (iii) Q =1 = Pno information )

Proof. We consider the case when Q<1,and need to show that if a 1/k Q, then the series k ! is absolutely convergent. The idea of the proof is essentially to mimic Example 5.72, with Q in place of 1/2. However, it turns that out we cannot use Q directly. The problem is that we could have the situation shown in the figure to the right. That 1/k is, even if ak converges to Q, this does not 1/k mean that ak Q has to hold for any value Fig. 22. Uh oh...  of k. 5.5. CONVERGENCE TESTS FOR POSITIVE SERIES 239

Fortunately, it is not too hard to get around this. If you choose your favourite number q so that Q1 to the reader (note that the divergence test could be useful).

Example 5.74 We use the root test to investigate whether the following series converges or not: 1 k2 . 2k Xk=1 To use the root test, we do the following computation:

2 2/k 2lnk/k 2limk ln k/k 0 k k k e e e 1 Q =lim =lim =lim = !1 = = . k 2k k 2 k 2 2 2 2 !1 r !1 !1 We conclude that the series converges since Q<1.

Exercise 5.75 Use the root test to check the convergence of

k2 1 k 1 1 1 1 (a) k (b) 1 (c) k kln k Xk=1 Xk=1 ✓ ◆ Xk=2 Exercise 5.76 In this exercise, we complete the proof of the root test for convergence.

(a) Fill in the details for the case Q>1. (b) Show that there exist both divergent and for which Q =1. Hint: Consider ↵-series. 240 CHAPTER 5. LIMITS FOR SEQUENCES

The quotient test

The quotient test is sort of the twin sibling of the root test. As with the root test, we start out by considering an example to convey the general idea of what is going on.

Example 5.77 Suppose that 1 a is a positive series so that for all k 1 we have k=1 k P a 1 k+1 . ak  2

But this is the same as a a k . k+1  2 Unlike for the root test, this does not immediately lead to a useful comparison. To see how to proceed, let us see what this means for the first few terms: a a 1 2  2 a a a 2 1 3  2  22 a a a a 3 2 1 4  2  22  23 Aha! In other words, we have a a 1 . k  2k k Since k1=1 a1/2 is a convergent geometric series, the standard comparison test implies that 1 a is convergent. Pk=1 k P The general formulation of the quotient test is as follows.

Proposition 5.78 (Quotient test for absolute convergence) Suppose that ak is a sequence of positive numbers, and that a lim k+1 = Q. k a !1 k Then the following hold:

(i) Q<1= a converges ) k (ii) Q>1= P a diverges ) k (iii) Q =1 = Pno information ) 5.5. CONVERGENCE TESTS FOR POSITIVE SERIES 241

Proof. The proof follows by modifying the example in the same way as for the root test. This time, let us go through the details in the case when Q>1. As in the case when Q<1,thelimita /a Q does not imply any inequality k+1 k ! between the terms ak+1/ak and Q. However, since Q>1, it follows that we will have a /a 1. But this is the same as having a a for k large enough. But k+1 k k+1 k this means that for k large enough, the sequence ak will be increasing. In particular, this implies that a 0. Hence, by the divergence test, the original series diverges. k 6!

Example 5.79 We now use the quotient test to investigate the series

1 k2 2k Xk=1 To this end, we make the following computation:

2 (k+1) 2 k ( 2k+1 ) (k + 1) 2 Q =lim 2 =lim k k k k2 · 2k+1 !1 ( 2k ) !1 k +1 2 1 1 1 =lim =1 = . k k · 2 · 2 2 !1 ⇣ ⌘ Since Q<1, we conclude that the series is convergent.

Exercise 5.80 Use the quotient rule to check whether the following series converge.

1 2k 1 1 1 1 (a) (b) 2 (c) k k k! k +1 e +e Xk=0 Xk=0 Xk=1 Exercise 5.81 In exercise 4.66, we briefly studied some infinite series representations for the functions ln(1 + x) and arctan(x). Here, we consider the following formulas: x2 x3 x4 ex =1+x + + + + 2 3! 4! ··· x3 x5 x7 sin(x)=x + + 3! 5! 7! ··· x2 x3 x4 ln(1 + x)=1 + + . 2 3 4 ··· (a) What does the quotient test say about each of the above series? (b) Verify numerically that each of the above formulas approximate their functions for x =1. For which of these series does the error estimate from the test allow you to predict how many terms to include for these approximations to have an error of at most 1/1000? 242 CHAPTER 5. LIMITS FOR SEQUENCES

5.6 Exam exercises

Exercises on the limit of sequences

Many exam exercises on limits of sequences also involve techniques that require knowl- edge on Taylor expansions. These exercises are delayed until the relevant chapter.

Exercise 5.82 (Lund, January 2016) Define what we mean by the symbol limn an n !1 and use this definition to show that limn ( 1) does not exist. !1 Exercise 5.83 (Lund, August 2015) (a) Give the definition for lim ak =0. k !1 (b) Use this definition to show that sin k lim =0. k p !1 k

Exercise 5.84 (Lund, May 2015) Give the definition of limn an =+ and use !1 1 it to show that n! lim = . n n !1 20 1

Exercises on the root and quotient tests for convergence

Exercise 5.85 (Exam 2015-08-17) Determine whether the following series con- verges. 1 2k . 1+lnk(k) Xk=2 Exercise 5.86 (Exam 2015-05-27, part of exercise) Determine whether the fol- low series converges or diverges. 1 k! . k2020k Xk=1 Exercise 5.87 (Exam 2014-12-18, part of exercise) Determine the convergence of 1 1 k2 1 . k Xk=1 ⇣ ⌘ Exercise 5.88 (Exam 2014-08-18, part of exercise) Determine the convergence of 1 k! . 3k Xk=1 5.6. EXAM EXERCISES 243

Exercise 5.89 (Exam 2013-05-29, part of exercise) Determine the convergence of 1 n3/2 . 5n ln(n) n=2 X 244 CHAPTER 5. LIMITS FOR SEQUENCES

5.7 Answers to selected exercises

5.4 (a) N =9and N = 13, respectively. (b) Use a while-loop that runs n=n+1 as long as 1/2 n > epsilon is true and then prints the final value of n. (c) Smaller ⇤⇤ ✏ leads to larger N. (d) For all ✏ 1/2, you can choose N =0. 5.5 (a) N =(5/✏ 3)/2, (b) N = 23,N = 234, (e) For all ✏ 1, you can choose N =0.

5.6 The key observation is that n! n. 5.8 (a) In the top figure, it seems clear that this is N =4, in the bottom, it is less clear, but N = 10 seems like a reasonable guess.

5.9 (a) For example, it works to apply the definition for positive sequences to a . n (b) Apply the definition for positive sequences to a . | n| 5.12 The key is to use the formula for partial sums of the Geometric series (see Chapter 4).

5.14 (a) an is not defined for n =3. This is not a problem as long as N is taken to be larger than or equal to 3.

5.16 (a) Given ✏, a suitable choice is N =(5/2)(3/✏ 1). (b) Here, the computation 2n 1 2n 2n 2 = <✏ n2 + n 1  n2 + n 1  n2 n works for n>N= max 1, 2/✏ (but other choices for N are also possible). { } 5.17 Use that 5n 5 5 5 5 = . n! 1 · 2 · 3 ···n 5.19 The limits are (a) 0, (b) 2, (c) 1/3.

5.22 (a) Given ✏,anyN works. (b) Given ✏, then N =1/✏ works.

5.23 The key observation is that for N large enough, n>N = a <✏/2 and )|n| b <✏/2. Combine this with the triangle inequality, and you are done. | n| 5.25 The limit is close to 2.71824....

5.26 (a) For instance, ⌦ > 0, N>0 such that for all n N we have n>N = 8 9 2 ) an > ⌦. (b) This is very similar to 5.23.

5.30 Here are the interpretations: (a) a = C + a . (b) C>0 and n !1 ) n !1 a = C a . n !1 ) · n !1 5.32 (a) 1, (b) 0. 5.7. ANSWERS TO SELECTED EXERCISES 245

5.33 (a) 2, (b) 1/2.

5.34 The third step is wrong.

5.35 It may be smart to divide this proof into two cases: (i) A =0and (ii) A>0.In the second case, the key observation is that x A px pA = . px + pA

5.37 (c) The limit should be 1.

5.38 The key is to observe that

n 1 n log(1+ 1 ) 1+ =e n . n ⇣ ⌘ 5.40 (a) a C/n C/n a C/n. (b) 0. | n| ()  n  5.41 For ↵>2 (then the limit is 0).

5.42 Yes (I will let you figure out an example).

5.43 In part (b),thelimitsare(i) 0, (ii) 1/2, (iii) . 1 5.47 (a) 1/5, (b) (5⇡ + 400)/105

5.49 In (a), (b) follow Example 5.48. To quickly solve (c), you can observe that bn bn n! = . nn n! · nn

5.50 (a) The inequality log n n 1 does the trick for all ↵>1. (c) The key is to  use, say, log(n)= 1 log(n2↵) 1 (n2↵ 1). 2↵  2↵ 5.51 Here, the key is to observe that

log n n e log n n log a = =e . an en log a

5.53 Choose ✏ = ✏ = L M /2 (or anything smaller than this). 1 2 | | 5.54 In part (a) the answers are (i) no, (ii) no, (iii) yes.

5.55 (b) ✏1 = ✏2 = ✏/2.

5.56 (b) ✏1 = ✏2 = ✏.

5.59 (a) Use the fact that for large enough n then a L 1,andthenapplytriangle | n | inequality to a after adding by 0 in a clever way. | n| 246 CHAPTER 5. LIMITS FOR SEQUENCES

5.60 (a) This follows by realising that a a + b + b is just the ordinary triangle | || | | | inequality in disguise. (b) Use the fact that for large enough n, then b M | n | M /2. After this, apply triangle inequality to b after adding by 0 in a clever | | | n| way.

5.61 (a) Well, L is the least upper bound of M,so...

5.64 Use k2 +1 k2 +0and pk2 + k +1 pk2 + k2 + k.  5.68 (a) Diverges (compare to harmonic series), (b) diverges (compare to a geometric series).

5.70 For L =0, = is valid, and for L = , = is valid. ( 1 ) 5.71 Converges if and only if ↵>1/2.

5.75 (a) converges, (b) converges, (c) no conclusion.

5.76 (a) This case is easier than Q<1. The reason is that from a 1/k Q>1,you | k| ! can deduce that a 1 for large enough k, and therefore the series diverges by the k divergence test. (b) Use the harmonic series and the Basel series as examples.

5.80 (a) converges, (b) no conclusion, (c) converges.