Math 123 - Shields Infinite Series Week 6
Zeno of Elea posed some philosophical problems in the 400s BC. His actual reasons for doing so are somewhat murky, but they are troubling regardless. In his Dichotomy paradox he argues that any objects in motion must arrive at the halfway point before they arrive at their goal. However, we can iterate the reasoning, and an object must also arrive at the halfway point to the halfway point first, and so on .... If the goal is m meters away and you travel at a speed of v, then the time needed to traverse the distance is x x x x t = + + + + ... 2v 4v 8v 16v x1 1 1 1 = + + + + ... v 2 4 8 16 The parenthetical expression is an infinite sum. Zeno said that the time for the motion then must be infinite since if we add infinitely many positive numbers together then this must result in a positive number. As beings of logic we are forced to conclude that motion is an illusion. We have all been duped.
On a completely unrelated note recall from Calc I that for small angles θ we can use a local linear approxi- mation to eθ. That is eθ ≈ 1 + θ This approximation is pretty good, but it’s not perfect because the approximation is a line and eθ bends. Well in a similar way we can make our estimate better by taking a quadratic approximation and say 1 eθ ≈ 1 + θ + θ2 2 This is better, but still ultimately an approximation with some built in error. However, we can keep making this more precise by adding on a cubed term, then a quadratic term, etc. What if we go off our rockers and decide to add infinitely many terms. Then maybe our approximation wouldn’t be an approximation at all, but it would become an identity? 1 1 1 eθ = 1 + θ + θ2 + θ3 + ... + θn + ... 2 6 n! Maybe that could make senes. But right now it doesn’t because we haven’t defined what an infinite series even means. The same thing holds for Mr. Zeno. He may have a point, but we need to be sure we have things defined before we start talking about properties that they may or may not have.
6.1 Sigma Notation & Convergence / Divergence
First we start by discussing finite sums. Consider the sum
1 + 2 + 3 + 4. It’s all well and good to write this out when the sum is short, but it would be frustrating if the sum were instead
S =1+2+3+4+5+ ... + 98 + 99 + 100. Instead we utilize sigma notation and write the above as
100 X S = n. n=1 The variable n is referred to as the index of the summation. The numbers 1 and 100 are called the bounds. The notation tells us to evaluate the expression inside the sum at all whole number values of the index n inclusively between the lower and upper numbers. More generally for example we have
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1 X 2an = a7 + a8 + a9 + a10 + a11 + a12. n=7 Notice that the indexing is somewhat arbitrary and can be manipulated. For example both expressions below define the same sum.
3 X 1 1 1 1 = 1 + + + 2n 2 4 8 n=0 5 X 1 1 1 1 = 1 + + + 2n−2 2 4 8 n=2
How do we bridge the gap to the infinite? Thinking about how we handled infinitely nested roots is produc- tive. An infinite sum is naturally realized as a sequence of finite sums which continually grow in length.
th Definition Let {an}n=i be a sequence. We define the k partial sum of the sequence as
i+k−1 X Sk = an = ai + ai+1 + ai+2 + ... + ai+k−2 + ai+k−2 n=i
th 1 Example The 4 partial sum of the sequence { n }n=1 is
4 X 1 1 1 1 S = = 1 + + + 4 n 2 3 4 n=1 Note: The kth partial sum will contain k terms.
∞ X n o Definition Let {an}n=i be a sequence. By the infinite sum an we mean the sequence Sk where k=1 n=i th Sk represents the k partial sum. If the sequence of partial sums converges, that is if lim Sk exists and is k→∞ equal to sum finite value S then we say that the sum converges to S. Otherwise we say that it diverges.
∞ X Example Does the series n converge? n=1 Examine the partial sums.
S1 = 1
S2 = 1 + 2 = 3
S3 = 1 + 2 + 3 = 6
S4 = 1 + 2 + 3 + 4 = 10 . . . n(n + 1) S = n 2 We have that lim Sn = ∞ so the series diverges. n→∞
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X 1 1 Example Does the series − converge? n n + 1 n=1 Again, look at the sequence of partial sums. 1 1 S = 1 − = 1 2 2 1 1 1 2 S = 1 − + − = 2 2 2 3 3 1 1 1 1 1 3 S = 1 − + − + − = 3 2 2 3 3 4 4
. . . 1 S = 1 − n n + 1
We have that lim Sn = 1. Therefore the series converges to 1. n→∞ A large part of the remainder of the course will be devoted to developing and applying tests to determine if series converge or diverge.
6.2 Divergence Test
Our first test gives us a condition which guarantees if a given series will diverge. It does not allows us to ever conclude that a series converges however.
∞ X Divergence Test The the series an converges then lim an = 0. In particular, if lim an 6= 0 then the n→∞ n→∞ n=i series diverges.
Proof For convenience assume that the series index begins at 1 and the series converges to the value S. In that case Sn = a1 + a2 + ... + +an−1 + an
Sn−1 = a1 + a2 + ... + +an−1 + an and an = Sn − Sn−1.
lim an = lim Sn − Sn−1 = S − S = 0 n→∞ n→∞
X n Example Does the series converge? n + 1 n=0 n lim an = lim = 1 6= 0 n→∞ n→∞ n + 1 so the series diverges.
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6.3 The Harmonic Series Diverges
As an important example of what the divergence test does not imply we have the harmonic series.
Definition The harmonic series is the series
∞ X 1 1 1 1 = 1 + + + + ... n 2 3 4 n=1
Theorem The harmonic series diverges.
Proof Look at the partial sums given by the powers of 2.
S1 = 1
1 S = S + 2 1 2 1 1 1 1 1 1 1 1 1 S = 1 + + + > 1 + + + = 1 + + 2 · = 1 + 2 · 4 2 3 4 2 4 4 2 4 2 1 1 1 1 1 1 S = S + + + + > S + 4 · > 1 + 3 · 8 3 5 6 7 8 3 8 2 Continuing this style of estimation we have that 1 S n > 1 + n · 2 2 1 This shows that the sequence Sn is not bounded above. Since Sn = Sn−1 + n , we can see that the sequence is also increasing. Therefore, it must diverge.
1 Note: lim = 0 and yet the harmonic series diverges. The divergent series test only gives us a condition n→∞ n for when a series will diverge. It cannot tell us that a series converges.
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6.4 Geometric Series
Definition A geometric series is any series which can be written in the form ∞ X arn i=0
Usually we do not care about the value at which our index begins. However, for geometric series we will assume that the series begin at 0. Geometric series are notably because, unlike most series we will encounter, they allow us to precisely determine the value of the sum.
∞ X Geometric Series Theorem The geometric series arn converges if |r| < 1 and otherwise diverges. The n=0 a value of the convergent sum is . 1 − r Proof 2 n Sn = a + ar + ar + ... + ar 2 n n+1 rSn = ar + ar + ... + ar + ar
a =⇒ S − rSn = a − arn+1 =⇒ S (1 − r) = a(1 − rn+1) =⇒ S = (1 − rn+1) n n n 1 − r n Now look at the limit, lim Sn. The only term dependent on n is the r . This converges if and only if n→∞ −1 < r ≤ 1 as we saw in the sequences section. However, the partial sum is not valid when r = 1. When r = 1, the divergence test shows us the series diverges. Therefore the series converges exactly when |r| < 1. With that assumption, taking the limit we have that a a S = lim Sn = (1 − 0) = n→∞ 1 − r 1 − r
Examples Determine if the following sums converge or diverge. If they converge, then find the value. ∞ X 1n (i) 2 i=0
1 1 This is geometric with a = 1 and r = 2 . |r| < 1 so the sum converges to 1 = 2. 1 − 2 ∞ X 3 (ii) 5n i=1 1 This is geometric with a = 3 and r = 5 . |r| < 1 so the series will converge. However, we cannot use the formula to determine its value. The formula assumes that the sum is indexed beginning at 0. We must first reindex.
∞ ∞ ∞ X 3 X 1 3 1 X 3 1 3 3 = = = · = 5n 5 5n−1 5 5n 5 1 − 1 4 i=1 i=1 i=0 5 ∞ X (iii) πne−n i=0 ∞ X π n This can be rewritten as . Therefore the series is geometric with a = 1 and r = π . Here e e i=0 |r| > 1 so the series diverges.
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(iv) Make sense of 0.999.
∞ X 9 We can view this repeating decimal as the sequence . After reindexing it can be written 10n i=1 ∞ X 1 9 . This is geometric with a = 9 and r = 1 . Therefore it converges to 10 10n 10 10 i=0
9 9 10 10 1 = 9 = 1 1 − 10 10
This is the origin of the unfortunate fact that decimal representations are not unique.
We refer to a series as telescoping when the partial sums of the series all have the same fixed number of non-zero terms due to cancellation. We can frequently determine the value of summation when a series is telescoping. We already saw one example of a series of this type in Example 3 from Section 6.1.
∞ X 2 Example The sum telescopes and converges to a value which we can determine. 2n(2n + 2) n=1 ∞ X 1 1 Using partial fractions the sum can be expanded to − . Looking at the partial sums, we can 2n 2n + 2 n=1 1 see that all terms cancel except the 2 and the smallest term in the partial sum. 1 1 S = − 1 2 4 1 1 1 1 1 1 S = − + − = − 2 2 4 4 6 2 6 1 1 1 1 S = S + − = − 3 2 6 8 2 8 1 1 1 1 S = S + − = − 4 3 8 10 2 10 1 1 Continuing we see that Sn = 2 − 2n+2 . Thus 1 1 1 1 lim Sn = lim − = − 0 = n→∞ n→∞ 2 2n + 2 2 2
∞ X 1 Example The sum telescopes and converges to a value which we can determine. n2 + 4n + 3 n=1 ∞ 1 X 1 1 Expanding by partial fractions you can write this as − . Now look at the partial sums. 2 n + 1 n + 3 n=1 1 1 S = − 1 2 4 1 1 1 1 S = − + − 2 2 4 3 5 1 1 1 1 1 1 S = S + − = + − − 3 2 4 6 2 3 5 6 1 1 1 1 1 1 S = S + − = + − − 4 3 5 7 2 3 6 7
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1 1 1 1 1 1 S = S + − = + − − 5 4 6 8 2 3 7 8 1 1 1 1 1 1 Continuing this pattern Sn = 2 + 3 − n+2 − n+3 . As n → ∞ we get that the sum converges to 2 + 3 .
∞ X n Example Be careful. Not all telescoping series converge. Consider the sum ln . n + 1 n=1 ∞ X Rewriting this as ln(n) − ln(n + 1) we can see that it will telescope. It has partial sums, n=1
S1 = ln(1) − ln(2)
S2 = S1 + ln(2) − ln(3) = ln(1) − ln(3)
S3 = S2 + ln(3) − ln(4) = ln(1) − ln(4)
Sn = ln(1) − ln(n + 1) = − ln(n + 1) Now lim − ln(n + 1) = −∞ so the series diverges. n→∞
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6.6 Properties of Convergent Series
The rub of it is that if a series converges, then it obeys the nice algebraic properties you would expect. Some results are collected.
∞ ∞ X X Theorem For j > i, an converges if and only if an converges. n=i n=j X X Theorem If an converges to A, bn converges to B, and c is a constant, then X 1. can converges to cA.
2. an ± bn converges to A ± B.
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