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 a 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 a partial sum of $Sigma a_k$. 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 $Sigma a_k$, 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.