Math 123- Shields Inﬁnite Sequences Week 3

3.2 Introduction

By a sequence we loosely mean an ordered list of numbers,

2, 4, 6, 8, 10, 12.

We refer to entries of the sequence as the terms of the sequence and to each term in the sequence we assign a positive whole number called its index. The index refers to the term’s place within the sequence. For example the above sequence could be written

a1, a2, a3, a4, a5, a6

where a1 = 2, a2 = 4, etc. The indices must be sequential, however, they need not start at 1. We could have instead notated the sequence by , b10, b11, b12, b13, b14, b15

where b10 = 2, b11 = 4, etc.

Sequences may also be inﬁnite in length. In fact we will only deal with inﬁnite sequences in this class so I will often drop the adjective when I speak. As an example of an inﬁnite sequence we could imagine a sequence which extends the above pattern on towards inﬁnity.

2, 4, 6, 8, 10, 12,...

The word pattern is problematic and must be made more rigorous. Sometimes people colloquially speak of a ”last term” in an inﬁnite sequence. This by itself is nonsense, but it can be given some real sort of meaning as we will see.

3.3 Inﬁnite Sequences & Convergence

Notation: Let N denote the collection of non-negative whole numbers: 0, 1, 2,... and N≥m the collection of non-negative whole numbers not smaller than m. For example N≥4 denotes the numbers 4, 5, 6,....

Deﬁnition A sequence of real numbers is a function f whose domain is N≥m and range is the real numbers.

This agrees with the intuitive idea of a sequence discussed in the introduction by taking an = f(n). For example, the sequence talked about there could be described by the function f(n) = 2n with domain N≥1. Although the formal deﬁnition is a function and it can be helpful to remember this, sometimes it is unnec- essary and we instead operate with the informal deﬁnition of a sequence as a list of numbers.

k Notation The ﬁnite sequence am, am+1, am+2, . . . , ak is denoted by {an}n=m and the inﬁnite sequence am, am+1, am+2,... by {an}n=m.

A sequence may deﬁned in any number of ways, but the salient point is that it must be done so unambigu- ously so that all parties agree on precisely what the terms of the sequence are. In this class we mainly do so in one of two ways.

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1) As an Explicit Function of n The most common way we will deﬁne a sequence is by giving an explicit function which allows us to compute the nth term of the sequence by simply evaluating the function.

Examples: 1 1 1 (i) { n }n=1 = 1, 2 , 3 ,...

1 From this we can easily compute any term directly. For example a100 = 100 . 1−n −1 −2 (ii) {e }n=0 = e, 1, e , e ,... n+1 n(−1) o 1 1 1 (iii) = 1, − 4 , 9 , − 16 ,... n2 n=1

n no (iv) an = 2 = 1, 2, 4, 8, 16, 32,... n=0

2) Recursively Occasionally it is more convenient to describe a sequence by giving some necessary initial values of the sequence and a formula to compute new values from the previous terms.

Examples:

(i) a0 = 1, a1 = 1, an = an−1 + an−2

Using this recursive formula we can compute as many terms of the sequence as is necessary. For example

a2 = a2−1 + a2−2 = a1 + a0 = 1 + 1 = 2 and a3 = a3−1 + a3−2 = a2 + a1 = 2 + 1 = 3 The sequence described is the well known Fibonacci Sequence. Note that with this method of deﬁning a sequence we cannot immediately compute an arbitrary value such as a100.

(ii) a0 = 1, an = 2an−1.

Here a1 = 2a0 = 2 · 1 = 2, a2 = 2a1 = 2 · 2 = 4, a3 = 2a2 = 2 · 4 = 8. Notice this gives the same n sequence as the one described by the formula an = 2 . 9 (iii) a = 0, a = a + . 0 n n−1 10n

This describes the sequence 0, 0.9, 0.99, 0.999, 0.9999, 0.99999,... A primary concern of ours will be the progression of sequences and whether their terms begin to settle down around a value or not. n o Deﬁnition A sequence an is said to converge if lim an exists and is ﬁnite. Otherwise it is said to diverge. n→∞ If lim an = L we will say that the sequence converges to L. n→∞

Real Deﬁnition which you can ignore but can also be useful if you take the time to understand n o it A sequence an is said to converge to L if for any real number > 0 there is some non-negative whole

number N such that for all n ≥ N we have that an − L < .

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Convergence sequence are nice in that they obey the expected arithmetic laws.

Theorem Let {an}, {bn} be sequences converging to a and b respectively and c a constant.

(i) {an ± bn} converges to a ± b.

(ii) {c · an} converges to c · a.

(iii) {an · bn} converges to a · b.

an a (iv) { } converges to b if b 6= 0. bn If we have a sequence given explicitly given as a function of n, then our primary technique to determine if a sequence converges or diverges will be to treat the sequence as a real function and use the techniques developed in Calc I to analyze it. This is justiﬁed by the following theorem.

Theorem Let {an} be a sequence and f(x) a function such that f(n) = an. Then if lim f(x) = L, then x→∞ lim an = L. n→∞ Examples:

n 1 o 1 (a) n converges to 0 since f(x) = x agrees with the sequence at all positive integer values and n=1 1 lim = 0 x→∞ x n o (b) n diverges since f(x) = x agrees with the sequence at all positive integer values and n=1 lim x = ∞ x→∞

n n o x (c) n+1 converges to 1. Take f(x) = x+1 . Then, n=0 x x 1 1 1 lim = lim · = lim = = 1 x→∞ x→∞ 1 x→∞ 1 x + 1 x 1 + x 1 + x 1 + 0

n n o x (d) en converges to 0. Take f(x) = ex . Then, n=0 x ∞ lim = x→∞ ex ∞ By L’Hopital’s Rule this limit is equivalent to 1 lim = 0 x→∞ ex

Note: Be careful. The theorem above only says what is says. It does not say that if lim f(x) does not x→∞ exists, the lim an does not exist. For example take an = sin(nπ). Then an = 0 for all n so the sequence x→∞ must converge to zero. However, taking f(x) = sin(nx) gives that lim f(x) does not exist as it oscillates x→∞ between −1 and 1.

The examples show the previous theorem to be a powerful tool in analyzing the convergence of sequences. Of course, it alone will be insuﬃcient to handle all sequences. Some other potentially useful theorems are documented here.

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n o Theorem Let an be a sequence.

n o (i) If lim an = 0, then an converges to 0. n→∞ n o (ii) If lim a2n = L and lim a2n+1 = L, then an converges to L. n→∞ n→∞ n o (iii) If lim a2n = L and lim a2n+1 = M where L 6= M, then an diverges. n→∞ n→∞ n o (iv) If bn ≤ an ≤ cn and lim bn = lim cn = L, then an converges to L. n→∞ n→∞ n o n o (v) If an converges to L, then any subsequence of an converges to L. n o (vi) If an converges to L and f(x) is continuous at L, then lim f(an) = f(L). n→∞

Examples: n(−1)n+1 o (i) converges to 0 since n n=1 (−1)n+1 1 lim = lim = 0 n→∞ n n→∞ n

n n+1 1 o (ii) π + (−1) n converges to π since n=1

2n+1 1 1 lim a2n = lim π + (−1) = lim π − = π − 0 = π n→∞ n→∞ n n→∞ n and 2n+2 1 1 lim a2n+1 = lim π + (−1) = lim π + = π + 0 = π n→∞ n→∞ n n→∞ n n 1 − no (iii) (−1)n diverges since n n=1

2n 1 − n 1 − n lim a2n = lim (−1) = lim = −1 n→∞ n→∞ n n→∞ n and 2n+1 1 − n 1 − n lim a2n+1 = lim (−1) = lim − = 1 n→∞ n→∞ n n→∞ n nsin(n)o −1 sin(n) 1 (iv) converges to 0 since ≤ ≤ and n n=1 n n n ±1 lim = 0 n→∞ n 1 (v) Assume that the sequence given by a0 = 0.1, an = 1 + converges to L. Knowing this we may 1 + an−1 determine L. √ 1 2 2 lim an = lim =⇒ L = 1 + =⇒ L + L = 1 + L + 1 =⇒ L = 2 =⇒ L = ± 2 n→∞ n→∞ 1 + L √ It does not take much more work to show that L = 2 since we begin with a positive number and can therefore only return positive numbers based on the recursive formula.

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n n sin 1 o 1 x (vi) e n converges to e since e is continuous at all real numbers and n=1

1 sin 1 sin n lim n sin = lim n = lim = 1 n→∞ n→∞ 1 n→0 n n n

Deﬁnition The factorial function n! is deﬁned for any non-negative integer n at the n − factor product

n! = n · (n − 1) · (n − 2) · ... · 2 · 1

Alternatively, we could deﬁne it recursively as 0! = 1, n! = n · (n − 1)!.

The ! is used for the factorial function to highlight the surprising rate at which is grows. For example 4! = 24 and 45! = 1.196 × 1056 which is approximately the number of atoms estimated to make up our solar system. n n! o Example Does converge? nn n=1

Examine a term an. This looks line n · (n − 1) ... 2 · 1 n − 1 2 1 = 1 · ... · n · n . . . n · n n n n

1 The quantity in parentheses is a product of numbers less than 1 and so itself less than 1. Therefore an ≤ n . Further all numbers are positive so 0 ≤ an. Since 1 lim = 0 = lim 0 n→∞ n n→∞ n n! o we can say that converges to 0. nn

n o Example For what values of r does rn converge? n=1

We may bound an on both sides,

n n −|r| ≤ an ≤ |r| Now ±|r|n converges to 0 if |r| < 1 and |r|n converges to 1 if r = 1. It will diverge otherwise. We may say that

0 |r| < 1 1 r = 1 lim rn = n→∞ Does not exist r ≤ −1 ∞ r > 1

Example What is the value of the inﬁnitely nested radical

s r q √ 90 + 90 + 90 + 90 + ... ?

The ﬁrst thing to note is that this question is utterly meaningless. What does the above expression even mean? Until I know what you are talking about, it is hopeless to answer your question. So, the ﬁrst step is to deﬁne what we mean by the value of an inﬁnitely nested radical. We will do this by realizing it as a sequence. If that sequence converges, we will call the value it converges to the value of the inﬁnitely

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n o √ √ nested radical. Consider the sequence an where a0 = 90 and an = 90 + an−1. Take a moment to n=0 appreciate that the limit of an is a very reasonable interpretation of the above expression. We will see later how to show that the sequence converges. For now, assume that it converges to L and ﬁnd its value.

√ p 2 2 lim an = lim 90 + an−1 =⇒ L = 90 + L =⇒ L = 90+L =⇒ L −L−90 = 0 =⇒ (L−10)(L+9) = 0 n→∞ n→∞ =⇒ L = 10 or L = −9 We can disregard the negative answer, so the value of the inﬁnitely nested radical is 10.

Example Consider the sequence given by the recursive relation an = 6 − an−1. Can we determine what a0 need to be ensure this converges? Can we determine what it converges to?

Why yes we can. By the recursive relation an−1 = 6 − an−2 so

an = 6 − (6 − an−2) = an−2

Therefore the sequence will bounce between two discrete values. For the sequence to converge we must ensure the values are the same. In other words we want a2 = a1 which implies that

a1 = 6 − a1 =⇒ 2a1 = 6 =⇒ a1 = 3.

It is then obvious that the sequence will converge to 3. Alternatively, you would see this by assume that the sequence converges to L and then solve

lim an = lim 6 − an−1 =⇒ L = 6 − L =⇒ 2L = 6 =⇒ L = 3. n→∞ n→∞

Note: It is not just a pedantic point that we cannot assume that the sequence converges before we ﬁnd a value. For example, take the sequence a0 = 1, an = (−1) · an−1. This sequence diverges, it just oscillates between the values −1 and 1. If we were to mistakenly forget to check this we would led to believe that 1 lim an = lim (−1) · an−1 =⇒ L = −L =⇒ 2L = 1 =⇒ L = n→∞ n→∞ 2

1 and the sequence converges to 2 . That would be a problem.

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