19  Convergence

Highlights of this Chapter: Finding the value of a series explicitly is difficult, so we develop some theory to determine convergence without explicitly finding the limit. Our main tool is comparison, which is built using the Monotone convergence theorem; and in particular comparison with a geometric series - the Ratio Test. Along the way to developing this theory we study a few important special series:

  • We prove the harmonic series 1n diverges.
  • In contrast, we prove that the sum of reciprocal squares 1n2 converges. In the final project we will show its value is π2/6.

In this section, we build up some technology to prove the convergence (and divergence) of series, without explicitly being able to compute the limit of partial sums. Such results will prove incredibly useful, as in the future we will encounter many theorems of the form if an converges, then… and we will need to a method of proving convergence to continue.

19.1 The Cauchy Criterion

For sequences, after some work we were able to find a definition equivalent to the original notion of convergence, which did not mention the precise value of the limit. This is exactly the sort of thing we seek for our investigation into series, so we carry it over directly here:

Definition 19.1 (Cauchy Criterion) A series sn=an satisfies the Cauchy criterion if for every ϵ>0 there is an N such that for any n,m>N we have |mnak|<ϵ

Exercise 19.1 Prove a series satisfies the Cauchy criterion if and only if its sequence of partial sums is a Cauchy sequence.

Because we know that being convergent and cauchy are equivalent, this means that all series that satisfy the Cauchy criterion are convergent, and conversely if a series does not, then it must diverge. We use this second observation to construct an easy-to-apply test for divergence:

Corollary 19.1 (Divergence Test) If a series an converges, then liman=0. Equivalently, if an0 then an diverges.

Proof. Let’s apply the cauchy condition to the single value m. This says for all ϵ>0 there is some N where for m>N we have |k=mmak|=|am|<ϵ

But making |am|<ϵ for all m>N is exactly the definition of am0.

This is useful mostly to immediately rule out the possibility that certain series converge. For instance it tells us that (1+1n) must diverge as the terms approach 1, not zero. But, when the terms approach zero its not very helpful: there are many series with an0 which do converge, and many which diverge. To distinguish between these, we need to build up some more powerful tools.

19.1.1 Absolute Convergence

Below we will develop several theorems that apply exclusively to series of positive terms. That may seem at first to be a significant obstacle, as many series involve both addition and subtraction! So, we take some time here to assuage such worries, and provide a means of probing a general series using information about its nonnegative counterpart.

Definition 19.2 (Absolute Convergence) A series an converges absolutely if the associated series of absolute values |an| is convergent.

Of course, such a definition is only useful if facts about the nonnegative series imply facts about the original. Happily, that is the case.

Theorem 19.1 (Absolute Convergence Implies Convergence) Every absolutely convergent series is a convergent series.

Proof. Let an be absolutely convergent. Then |an| converges, and its partial sums satisfy the Cauchy criterion. This means for any ϵ we can find an N where |an|+|an+1|++|am|<ϵ

But, by the triangle inequality we know that |an+an+1++an||an|+|an+1|++|am| Thus, our original series ak satisfies the Cauchy Criterion, as |k=mnak|<ϵ And, since Cauchy is equivalent to convergence, this implies ak is a convergent series.

19.2 Comparison

One of the very most useful convergence tests for a series is comparison. This lets us show that a series we care about (that may be hard to compute with) converges or diverges by comparing it to a simpler series - much like the squeeze theorem did for us with sequences. This theorem gives less information than the squeeze theorem (it doesn’t give us the exact value of the series we are interested in) but it is also easier to use (it only requires a bound, not an upper and lower bound with the same limit).

Theorem 19.2 (Comparison For Series) Let an and bn be two series of nonnegative terms, with 0anbn.

  • If bn converges, then an converges.
  • If an diverges, then bn diverges.

The proof is just a rehashing of our old friend, Monotone Convergence.

Proof. We prove the first of the two claims, and leave the second as an exercise. If xn0 then the series sn=k=0nxk is monotone increasing (as by definition sn=sn1+xn and xn0 we see snsn1 for all n).

Thus. an and bn are monotone sequences. If bn converges, we know by the Monotone Convergence Theorem that it its limit β is the supremum of the partial sums, so for all n k=0nbkβ But, since akbk for all k, we see the same is true of the partial sums k=0nakk=0nbk Stringing these inequalities together, we see that ak is bounded above by β. Since it is monotone (as the sum of nonnegative terms) as well, Monotone convergence assures us that it converges, as claimed.

Exercise 19.2 Let an and bn be two series of nonnegative terms, with 0anbn. Prove that if an diverges, then bn diverges.

The comparison test is incredibly useful: two of the most famous series it lets us understand are left as exercises below.

Exercise 19.3 Prove that 1n2 converges. *Hint: compare with 1/((n1)n), which telescopes.

Exercise 19.4 Show the harmonic series 1n diverges, by comparing it with the partial sums of 1,1/2,1/4,1/4,1/8,1/8,1/8,1/8,1/16,...

19.3 The Ratio Test

We saw in the last chapter that geometric series - where the consecutive ratios of every pair of terms is constant - are particularly easy to sum. Now that we have comparison, we can leverage this to provide a powerful convergence test for a much larger collection of series: those whose consecutive rations are constant in the limit.

Theorem 19.3 (The Ratio Test) Let an be a series, and assume that the sequence of consecutive ratios converges, lim|an+1an|=α Then an converges if α<1, and diverges if α>1.

Proof. We prove the convergence claim for α<1 here, and leave the divergence for α>1 as an exercise.

Assume that lim|an+1an|<1, and let N be such that for all n>N we have |an+1an|<r For some fixed 0<r<1 (perhaps do this by choosing your favorite ϵ>0, defining r=1ϵ and using the convergence hypothesis). Thus, for all n>N we know |an+1|<r|an|, and so inductively |aN+n|<rn|aN|. Summing the series, we see that k=0n|aN+k|<k=0nrk|aN| Thus, starting from the Nth term, our series is bounded above by a multiple of a geometric series! And, since we know geometric series converge, we can use comparison to see that kN|ak| is convergent.

But the first finitely many terms of a series cannot affect whether or not it converges, so we see that

k0|ak| is convergent

This is the definition of ak being absolutely convergent, and thus ak is itself convergent.

Exercise 19.5 Prove that if lim|an+1an|>1, the series an diverges.

Note that this test does not tell us anything when α=1: it only says that our series is growing faster than a geometric series with r<1 but slower than such a series with r>1. There is plenty of room for both behaviors in this gap:

Example 19.1 The sequence 1n diverges, but its limiting ratio is lim|1n+11n|=lim|nn+1|=1

But, the sequence 1n2 converges, with the same limiting ratio: lim|1(n+1)21n2|=lim|n2(n+1)2|=1

Remark 19.1. There is an even more general version of the ratio test were we don’t assume that |an+1/an| converges, but only that it is eventually strictly bounded above by 1. Precisely, all that’s actually required is limNsup{|an+1an|nN}<1

Exercise 19.6 Prove that the following series converges: n01n!

19.4 Other Convergence Tests

Because series are ubiquitous throughout mathematics, there are many more convergence theorems that have been developed than we have the time to cover here. Though we will not need them in our course, I list two of the most popular (following the ratio test) below for reference.

Theorem 19.4 (The Root Test) Let an be a series, and assume that the sequence of nth roots converges, lim|an|n=α Then an converges if α<1, and diverges if α>1.

The following test shows up in a Calculus II course; though we are not ready to rigorously discuss it yet as it requires integration. Once we gain some ability with integrals, this will allow us to leverage our abilities with the Fundamental Theorem to prove new facts about series.

Theorem 19.5 (The Integral Test) If f is a continuous function such that the sequence an=f(n) is a defined by evaluating f at integer values, then the sum an converges if and only if the integral 0f(x)dx converges.

19.5 Conditionally Convergent Series

Definition 19.3 A series converges conditionally if it converges, but is not absolutely convergent.

Such series caused much trouble in the foundations of analysis, as they can exhibit rather strange behavior. We met one such series in the introduction, the alternating sum of 1/n which seemed to converge to different values depending on the order we added its terms. Here we begin an investigation into such phenomena.

19.5.1 Alternating Series

Definition 19.4 (Alternating Series) An alternating series is a series of the form (1)nbn for an a nonnegative series. That is, every term switches from positive to negative.

Theorem 19.6 (Alternating Series Test) If (1)nan is alternating, then it converges if an decreases monotonically with limit zero.

Before jumping in, its helpful to take a look at a few partial sums to start. For example, s4:

s4=a0a1+a2a3+a4=(a0a1)+(a2a3)+a4

Grouping the terms of this finite sum like so shows that s4 is a sum of positive numbers (since an is decreasing, so anan10): thus s40.

s4=a0a1+a2a3+a4=a0(a1a2)(a3a4) This grouping shows s4 is equal to a0 minus a bunch of nonnegative terms: thus s4a0. This extends directly

Exercise 19.7 Let sn=k=0n(1)kak be an alternating series with an0 monotonically. Prove by induction that

  • All the partial sums sn are nonnegative.
  • All partial sums are bounded above by the first term a0.

Corollary 19.2 Starting the sum at N instead of 0, the same argument shows that |k=Nn(1)kak||aN| for all nN.

What other patterns can we notice? Increasing from s4 to s6 we see s6=a0a1+a2a3+a4a5+a6 =s4a5+a6=s4(a5a6) Thus s6s4. A similar look at s3 and s5 shows s5=a0a1+a2a3+a4a5=s3+(a4a5) So s5s3! This is a sort of pattern we’ve seen before, where it’s helpful to look at the even versus odd subsequences individually:

Exercise 19.8 Let sn=k=0n(1)kak be an alternating series, and prove by induction that

  • The even subsequence is monotone decreasing
  • The odd subsequence is monotone increasing

Because each of these subsequences is monotone and bounded (by the previous exercise) they converge via monotone convergence. Now, all we need to see is they converge to the same limit to assure convergence of the entire series, by .

Proposition 19.1 Let sn=k=0n(1)kak be an alternating series with an0 monotonically. Then sn converges.

Proof. Let en=s2n and on=s2n+1 be the even and odd subsequences respectively, and note that on=ena2n+1. Then, since we know the subsequence a2n+1 converges to zero (as an0, so all subsequences have the same limit) we can apply the limit theorems and see limon=limena2n+1=limenlima2n+1=limen So, the odd and even subsequences do have the same limit, as required.

19.5.2 Properties of Conditionally Convergent Series

First we look at the main example of a conditionally convergent series.

Example 19.2 (1)nn is conditionally convergent:

  • It converges, by the alternating series test.
  • But it is not absolutely convergent, as 1n diverges by EXR

This series is famous from the introduction to our course, where we saw that its value when summed is the natural logarithm of 2, but that this value changes when we reorder the terms! This is a general behavior of conditionally convergent series; and one hint of this is that the sum of their positive and negative terms separately each diverges to ±.

Theorem 19.7 If ak is conditionally convergent, let pk be the subsequence of all positive terms of ak and nk be the subsequence of all negative terms. Prove that pknk

For an absolutely convergent series, this cannot happen, and the sums of all the positive terms converges, as does the sum of all the negative terms.

Exercise 19.9 Prove that if an is absolutely convergent, then its subseries of positive terms and its subseries of negative terms both converge.