More elegant proof for $\sum \frac{a_n}{s_n}$ diverges.

Question

Here is the question I am trying to proof:

Suppose $a_n > 0$ for each $n$ and $\sum a_n$ diverges. Let $s_n = a_1 + \dots + a_n.$

$(a)$ Prove that $$\frac{a_{N+1}}{s_{N + 1}} + \frac{a_{N+2}}{s_{N + 2}} + \dots + \frac{a_{N+k}}{s_{N + k}} \geq 1 - \frac{s_N}{s_{N + k}} \quad \quad \forall N \geq 1, k \geq 1.$$

$(b)$ Deduce that $\sum \frac{a_n}{s_n}$ diverges.

My question is:

I managed to prove part (a), and here is a trial for part (b):

My question is:

Is there a more elegant way of proving part(b)? Could someone help me answer this question please?

Answer 1

A solution that is not much different than yours, but perhaps a bit more succinct, is to say that for fixed $N\in \mathbb{N}$ and for every $M\geq N$ from part (a) you have $$\sum_{n=N+1}^M \frac{a_n}{s_n} \geq 1 -\frac{s_N}{s_M}$$ and letting $M\to\infty$ deduce that for all $N\in \mathbb{N}$ $$\sum_{n=N+1}^\infty\frac{a_n}{s_n}\geq 1$$ This is a contradiction because the tails of convergent series always converge to $0$.