Advanced Calculus Single Variable

5.1.1 Absolute Convergence

Proof: Let ε > 0 be given. Then by assumption and Theorem 5.1.7, there exists n_ε such that whenever q ≥ p ≥ n_ε,

∑q |ak| < ε. k=p

Therefore, from the triangle inequality,

| | ∑q ||∑q || ε > |ak| ≥ || ak||. k=p |k=p |

By Theorem 5.1.7, ∑ _k=m^∞a_k converges. ■

In fact, the above theorem is really another version of the completeness axiom. Thus its validity implies completeness. You might try to show this.

One of the interesting things about absolutely convergent series is that you can “add them up” in any order and you will always get the same thing. This is the meaning of the following theorem. Of course there is no problem when you are dealing with finite sums thanks to the commutative law of addition. However, when you have infinite sums strange and wonderful things can happen because these involve a limit.

Proof: From absolute convergence, there exists M such that

∑∞ |ak| < ε k=M+1

Since θ is one to one and onto, there exists N ≥ M such that

⊆. It follows that it is also the case that

∑∞ || || aθ(k) < ε k=N+1

This is because the partial sums of the above series are each dominated by a partial sum for ∑ _k=M+1^∞

since every index θ equals some n for n ≥ M + 1. Then since ε is arbitrary, this shows that the partial sums of ∑ aθ

(k)

are Cauchy. Hence, this series does converge and also

||M N || ∞ ||∑ a − ∑ a ||≤ ∑ |a | < ε |k=1 k k=1 θ(k)| k=M+1 k

Hence

||∞∑ ∞∑ || ||∞∑ ∑M || ||M∑ ∑N || || ak − aθ(k)||≤ || ak − ak||+ || ak − aθ(k)|| |k=1 k=1 | |k=1 k=1 | |k=1 k=1 |

| | ||N∑ ∑∞ || ∑∞ ∞∑ | | + || aθ(k) − aθ(k)|| < |ak|+ ε + |aθ(k)| < 3ε k=1 k=1 k=M+1 k=N+1

Since ε is arbitrary, this shows the two series are equal as claimed. ■

So what happens when series converge only conditionally?

First of all consider why it converges. Notice that if S_n denotes the n^th partial sum, then

Thus the even partial sums are decreasing and the odd partial sums are increasing. The even partial sums are bounded below also. (Why?) Therefore, the limit of the even partial sums exists. However, it must be the same as the limit of the odd partial sums because of the last equality above. Thus lim_n→∞S_n exists and so the series converges. Now I will show later below that ∑ _k and ∑ _k both diverge. Include enough even terms for the sum to exceed 7. Next add in enough odd terms so that the result will be less than 7. Next add enough even terms to exceed 7 and continue doing this. Since 1∕k converges to 0, this rearrangement of the series must converge to 7. Of course you could also have picked 5 or −8 just as well. In fact, given any number, there is a rearrangement of this series which converges to this number.

Proof: Consider the first claim. From the assumption, there exists n^∗ such that n^∗ > max

and for all n ≥ n^∗ b_n ≥ a_n. Then if p ≥ n^∗,

Thus the sequence,_p=m^∞ is bounded above and increasing. Therefore, it converges by completeness. The second claim is left as an exercise. ■

For n > 1,

1 1 n2 ≤ n-(n−-1).

Now

Therefore, letting a_n = and b_n =

A convenient way to implement the comparison test is to use the limit comparison test. This is considered next.

Proof:Let n^∗ be such that n ≥ n^∗, then

an an ---> a and---< b bn bn

and so for all such n,

ab < a < bb n n n

and so the conclusion follows from the comparison test. ■

The following corollary follows right away from the definition of the limit.

This series converges by the limit comparison test above. Compare with the series of Example 5.1.13.

Therefore, the series converges with the series of Example 5.1.13. How did I know what to compare with? I noticed that is essentially like = n² for large enough n. You see, the higher order term n⁴ dominates the other terms in n⁴ + 2n + 7. Therefore, reasoning that 1∕ is a lot like 1∕n² for large n, it was easy to see what to compare with. Of course this is not always easy and there is room for acquiring skill through practice.

To really exploit this limit comparison test, it is desirable to get lots of examples of series, some which converge and some which do not. The tool for obtaining these examples here will be the following wonderful theorem known as the Cauchy condensation test.

Proof:This follows from the inequality of the following claim.

Claim:

∑n ∑2n n∑ 2ka2k−1 ≥ ak ≥ 2k− 1a2k. k=1 k=1 k=0

Proof of the Claim: Note the claim is true for n = 1. Suppose the claim is true for n. Then, since 2ⁿ⁺¹ − 2ⁿ = 2ⁿ, and the terms, a_n, are decreasing,

n+1 n 2n ∑ 2ka k−1 = 2n+1a n +∑ 2ka k−1 ≥ 2n+1a n + ∑ a k=1 2 2 k=1 2 2 k=1 k

2n+1 2n n n+1 ≥ ∑ a ≥ 2na n+1 + ∑ a ≥ 2na n+1 + ∑ 2k−1a k = ∑ 2k− 1a k.■ k=1 k 2 k=1 k 2 k=0 2 k=0 2

Let a_n =

. Then a_2ⁿ = ⁿ. From the Cauchy condensation test the two series

∞∑ ∑∞ ( )n ∑∞ ( )n -1-and 2n -1 = 2(1−p) n=1np n=0 2p n=0

converge or diverge together. If p > 1, the last series above is a geometric series having common ratio less than 1 and so it converges. If p ≤ 1, it is still a geometric series but in this case the common ratio is either 1 or greater than 1 so the series diverges. It follows that the p series converges if p > 1 and diverges if p ≤ 1. In particular, ∑ _n=1^∞n⁻¹ diverges while ∑ _n=1^∞n⁻² converges.

Use the limit comparison test.

(1) lnim→∞ (----n1---)-= 1 √n2+100n-

and so this series diverges with ∑ _k=1^∞

.

Sometimes it is good to be able to say a series does not converge. The n^th term test gives such a condition which is sufficient for this. It is really a corollary of Theorem 5.1.7.

Proof:Apply Theorem 5.1.7 to conclude that

n lim a = lim ∑ a = 0.■ n→∞ n n→∞ k=n k

It is very important to observe that this theorem goes only in one direction. That is, you cannot conclude the series converges if lim_n→∞a_n = 0. If this happens, you don’t know anything from this information. Recall lim_n→∞n⁻¹ = 0 but ∑ _n=1^∞n⁻¹ diverges. The following picture is descriptive of the situation.