Series convergence question from baby Rudin. 
Suppose, $a_n>0$, $s_n=a_1+\cdots +a_n$, and $\sum a_n$ diverges.  Prove that
  $$\frac{a_{N+1}}{s_{N+1}}+\cdots\frac{a_{N+k}}{s_{N+k}}\ge 1-\frac{s_N}{s_{N+k}}$$
  and deduce that $\sum\frac{a_n}{s_n}$ diverges.

Looking for a verification of my proof.
$$\begin{align}
\sum_{m=1}^k\frac{a_{N+m}}{s_{N+m}} &\ge\frac{1}{s_{N+k}}\sum_{m=1}^ka_{N+m}\\
& ={s_{N+k}-s_N\over s_{N+k}}\\
& = 1-\frac{s_N}{s_{N+k}}.
\end{align}$$
Let $0<\epsilon<1$. Since, $\lim_{k\to\infty}1-\frac{s_N}{s_{N+k}}=1$,
there is a $K\ge 1$, such that $k\ge K\implies 1-\left(1-\frac{s_N}{s_{N+k}}\right)<\epsilon$; that is $\frac{s_N}{s_{N+k}}<\epsilon$. So, it has been shown that there is an $\epsilon>0$ such that for all $N\ge 1$,
$\sum_{m=1}^k\frac{a_{N+m}}{s_{N+m}}>1-\epsilon$. Therefore the series does not meet the Cauchy criterion and diverges.
 A: Here may be an intuitive way to think about it. Assume the sequence is convergent, I claim that there exist some $\epsilon$ such that for all $N>0$, there exists terms 
$$
\sum^{n}_{i=m}\frac{a_{n}}{s_{n}}\ge \epsilon, n>m> N
$$
Therefore the sequence is not convergent and it must diverge. To see the claim I made, since we know $s_{n}\rightarrow \infty$ as $n\rightarrow \infty$, for any sufficiently large $N$, we can decompose $s_{n}=s_{N}+\sum^{N}_{i=n+1}a_{i}$. Since the $s_{N}$ factor is going to be relatively small in contrast to $s_{n}$ with $n\rightarrow \infty$, it is enough to choose enough factors from the sum to get $$\sum^{n}_{i=m}\frac{a_{i}}{s_{i}}\ge \frac{\sum^{n}_{i=m}a_{i}}{s_{n}}\ge \epsilon$$
And I think this is essentially what Rudin's hint and your proof is about. Of course this kind of result would not hold if $\sum a_{i}$ converges, as the sum would converge as well. 
A: Your proof that $$\sum_{m=1}^k \frac{a_{N+m}}{s_{N+m}} \geq 1 - \frac{s_N}{s_{N+k}}$$ looks correct, except that I think, technically, since $a_n > 0$ and hence $s_{n+1} > s_n$ for all $n$, $$\sum_{m=1}^k \frac{a_{N+m}}{s_{N+m}} > \frac{1}{s_{N+k}} \sum_{m=1}^k a_{N+m}.$$  So strictly greater with no chance for equality.  You also may want to remark the (slightly obvious point) that $s_{n+1} > s_n$ beforehand, for completeness.
As for the second part of the proof, maybe you could be a bit more explicit and say why the series does not meet the Cauchy criterion?  I.e. you could say that an $\epsilon > 0$ cannot be chosen arbitrarily due to what you've shown.
In other words, your argument looks good and I'm only saying that you could (optional!) explain your steps in a more complete manner.
