# Comparison Test Proof (Hammack's Book of Proof, Third Edition, Exercise 13.8.2)

I am currently going through Hammack's Book of Proof (3rd edition), and have been stuck on the following exercise for the past two weeks.

Prove the comparison test: Suppose $$\Sigma a_k$$ and $$\Sigma b_k$$ are series. If $$0 \le a_k \le b_k$$ for each $$k$$, and $$\Sigma b_k$$ converges, then $$\Sigma a_k$$ converges. Also, if $$0 \le b_k \le a_k$$ for each $$k$$, and $$\Sigma b_k$$ diverges, then $$\Sigma a_k$$ diverges.

Now, the reader is instructed to prove this using "Definition 13.7 (and Definition 13.5, as needed)" from the book which goes as follows.

Definition 13.7

A series $$\sum_{k=1}^{\infty} a_k$$ converges to a real number $$S$$ if its sequence of partial sums $$\{s_n\}$$ converges to $$S$$. In this case we say $$\sum_{k=1}^{\infty} a_k = S$$.

We say $$\sum_{k=1}^{\infty} a_k$$ diverges if the sequence $$\{s_n\}$$ diverges. In this case $$\sum_{k=1}^{\infty} a_k$$ does not make sense as a sum or does not sum to a finite number.

Definition 13.5

A sequence $$\{a_n\}$$ converges to a number $$L \in \mathbb{R}$$ provided that for any $$\epsilon > 0$$ there is an $$N > \mathbb{N}$$ for which $$n > N$$ implies $$|a_n - L| < \epsilon$$.

If $$\{a_n\}$$ converges to $$L$$, we denote this state of affair as $$\lim_{n \to \infty} a_n = L$$

If $$\{a_n\}$$ does not converge to any number $$L$$, then we say it diverges.

I am currently focusing on the first statement. Also, that statement intuitively makes sense to me: if the series $$\Sigma a_k$$ has only non-negative terms, then it will either converge or diverge to infinity. However, since $$0 \le a_k \le b_k$$, it follows that $$\Sigma a_k \le \Sigma b_k$$ and thus, if $$\Sigma a_k$$ did diverge to infinity we would have a contradiction. I am able to formalize this as follows.

Proof. For the sake of contradiction, suppose $$\Sigma a_k$$ diverges to infinity. That is, for every $$L \in \mathbb{R}$$ there exists a number $$N' \in \mathbb{N}$$ for which $$n \ge N'$$ implies $$A_n > L$$ where $$A_n$$ is the $$n$$-th partial sum of $$\Sigma a_k$$. Also, because $$\Sigma b_k$$ converges to some number $$B \in \mathbb{R}$$, we have that for every $$\epsilon > 0$$, there exists a number $$N'' \in \mathbb{N}$$ for which $$n > N''$$ implies $$|B_n - B| < \epsilon$$ where $$B_n$$ is the $$n$$-th partial sum of $$\Sigma b_k$$. Further, notice that if $$0 \le a_k \le b_k$$ for each $$k$$, then $$A_n = (a_1 + a_2 + \dots + a_n) \le (b_1 + b_2 + \dots + b_n) = B_n$$.

Now, take $$\epsilon > 0$$ and $$L = B - \epsilon$$. Also, let $$N = \max(N', N'')$$. Then, if $$n > N$$ we have $$A_n > B - \epsilon$$ and $$|B_n - B| < \epsilon$$, or $$B_n < \epsilon + B$$. Thus we have $$B_n < \epsilon + B < A_n$$ and $$A_n \le B_n$$, a contradiction. Therefore, $$\Sigma a_k$$ must converge. $$\tag*{\blacksquare}$$

The problem I have is that although divergence to infinity implies divergence, divergence does not imply divergence to infinity. Therefore, for my proof to be complete, I would need to show that if $$\Sigma a_k$$ diverges then it must diverge to infinity. However, I am unable to formalize this idea. I have looked it up online and stumbled accross proofs that uses Cauchy sequences or the monotone convergence theorem, but nothing that uses only the definitions above. I also tried a direct proof without any luck.

Any pointers or help would be greatly appreciated.

• If $\sum a_k$ does not converge, then $\sum a_k$ does not converge to $B$. The definition of divergence, then (any divergence, not just infinity) tells us that there is an $\epsilon>0$ such that for every $N>0$ there is an $n>N$ such that $|A_n-B|>\epsilon$. You can use that to prove your contradiction. Mar 23, 2021 at 15:17
• @c_gnar , so from $|A_n - B| > \epsilon$ you have that $A_n - B > \epsilon$ or $A_n - B < - \epsilon$ right? If so, the former leads to a contradiction since from $B_n - B < \epsilon$ it follows that $B_n - B < A_n - B$, and thus $B_n < A_n$, a contradiction. However, I fail to see how to get to a contradiction from the later. Mar 23, 2021 at 15:42
• You're right, I made a mistake. Try this: $A_n$ is a non-decreasing sequence. Therefore, if $A_n$ is bounded above, it must converge (least upper-bound principle). That should take care of the divergence $\Rightarrow\to\infty$ issue you raised. Mar 23, 2021 at 18:44
• Ok I see. Well, you can for sure start with the following: Letting $s_n=\sum_{k=1}^na_k$, $s_k\leq s_{k+1}$. That is easy to prove directly. This rules out oscillating sequences like the one you proposed. Truthfully, I don't know but I'm doubtful that it's possible to prove non-decreasing sequences converge or to go infinity without using the least upper-bound principle. It is a well-known theorem called the monotone convergence theorem. Mar 24, 2021 at 20:27
• That's why I suggested more clarity on the book's definition of divergence. I was thinking perhaps they add some details that either state or imply divergence to infinity or nonsense sum. Mar 24, 2021 at 20:29

I finally managed to complete my proof by showing that if $$\Sigma a_k$$ diverges, then it must diverge to infinity.
Proof. Suppose the series $$\Sigma a_k$$ diverges. Then, for all $$L \in \mathbb{R}$$, there exists a $$\epsilon > 0$$ such that for all $$N \in \mathbb{N}$$, there is a number $$n>N$$ for which $$|A_n - L| \ge \epsilon$$ where $$A_n$$ is the $$n$$-th partial sum of $$\Sigma a_k$$. Also suppose that $$a_k \ge 0$$ for all $$k$$. We must show that this implies that $$\Sigma a_k$$ diverges to infinity, or, in other terms, that for all $$L > 0$$, there is a $$N \in \mathbb{N}$$ for which $$n > N$$ implies $$A_n > L$$.
Now, from $$|A_n - L| \ge \epsilon$$ we have $$(A_n - L \ge \epsilon) \lor (A_n - L \le - \epsilon)$$.
Case 1. Suppose $$A_n - L \ge \epsilon$$. Thus, we have a number $$\epsilon > 0$$ such that for all $$N \in \mathbb{N}$$ there is a $$n > N$$ for which $$A_n \ge L+ \epsilon$$. However, from $$a_k \ge 0$$ we have that $$A_{n+1} \ge A_n$$ for all $$n$$. Therefore, $$n' > n$$ implies $$A_{n'} \ge A_n \ge \epsilon + L$$. That is, for all numbers $$L+\epsilon = L' > 0$$, we have a $$N \in \mathbb{N}$$ for which $$n > N$$ implies $$A_n > L'$$. That is, by definition, $$\Sigma a_k$$ diverges to infinity.
Case 2. Suppose $$A_n - L \le -\epsilon$$. Now, from the definition of divergence, we also have a $$N' > n$$ for which there is a number $$n' > N'$$ such that $$|A_{n'} - L| \ge \epsilon$$, or $$(A_{n'} - L \ge \epsilon) \lor (A_{n'} - L \le -\epsilon)$$. From case 1 above we know that $$A_{n'} - L \ge \epsilon$$ implies convergence to infinity. Now, for $$A_{n'} - L \le -\epsilon$$, taking the difference of $$A_{n'}$$ and $$A_n$$ we get $$A_{n'} - A_{n} \le 0$$. That is, $$(a_{n+1} + a_{n+2} + \dots + a_{n'}) \le 0$$ and so it follow that $$a_{n'} = 0$$ for all $$n' > N' > n$$. Then, for any $$\epsilon > 0$$, we have a number $$n \in \mathbb{N}$$ such that $$n' > n$$ implies $$|A_{n'} - A_n| = 0 < \epsilon$$. Therefore, by definition, we have that $$\Sigma a_k$$ converges, a contradiction.
The above two cases show that if $$\Sigma a_k$$ diverges, then it must diverge to infinity. $$\tag*{\blacksquare}$$