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.

  • $\begingroup$ 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. $\endgroup$
    – c_gnar
    Mar 23, 2021 at 15:17
  • $\begingroup$ @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. $\endgroup$ Mar 23, 2021 at 15:42
  • $\begingroup$ 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. $\endgroup$
    – c_gnar
    Mar 23, 2021 at 18:44
  • 1
    $\begingroup$ 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. $\endgroup$
    – c_gnar
    Mar 24, 2021 at 20:27
  • 1
    $\begingroup$ 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. $\endgroup$
    – c_gnar
    Mar 24, 2021 at 20:29

1 Answer 1


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. $$\tag*{$\blacksquare$}$$


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