Proving that $\sum\limits_{n=1}^∞\frac{a_n}{s_n^2}$ converges 
Let $a_n>0\;$ ($n=1,2,...\,$) with $\sum\limits_{n=1}^\infty a_n$ divergent and $s_n =\sum\limits_{k=1}^na_k$. For all $n \ge 2$,  prove that $\sum\limits_{n=1}^∞\dfrac{a_n}{s_n^2}$ converges. 

Proof:  For all $n \ge 2$, we have $\dfrac{a_n}{s_n^2} \le \dfrac1{s_{n-1}} -\dfrac1{s_n}$ and $\sum\limits_{n=2}^{\infty} \dfrac{a_n}{s_n^2} \le \sum\limits_{n=2}^{k} \left(\dfrac{1}{s_{n-1}} - \dfrac{1}{s_n} \right)$.
Now $\sum\limits_{n=2}^k \left (\dfrac{1}{s_{n-1}} - \dfrac{1}{s_n} \right) = \dfrac{1}{s_1} - \dfrac{1}{s_k}$ converges to $\dfrac{1}{s_1}$. This follows because $\sum\limits_{n=1}^\infty a_n$ diverges and $\dfrac{1}{s_n} \to 0$ as $ n \to \infty$. Thus, by the comparison test, $\sum\limits_{n=1}^{\infty} \dfrac{a_n}{s_n^2}$ converges.
Is this proof  correct?
$\dfrac{a_n}{s_n^2} \le \dfrac1{s_{n-1}} -\dfrac1{s_n}$
Proof: Let $n \le 2$
$ s_{n-1} \le s_{n}$ $\Leftrightarrow \frac{1}{s_{n}} \le \frac{1}{s_{n-1}} $
$ \Leftrightarrow \frac{1}{s_{n^2}} \le \frac{1}{s_{n}s_{n-1}}$
$ \Leftrightarrow   \frac{a_{n}}{s_{n^2}} \le  \frac{a_{n}}{s_{n}s_{n-1}} = \frac{s_{n} - s_{n-1}}{s_{n}s_{n-1}}$
$\Leftrightarrow  \frac{a_{n}}{s_{n^2}} \le \frac{1}{s_{n-1}} - \frac{1}{s_{n}} $

Your proof is correct if $a_n$ is nonnegative since it then follows that
$$\frac{a_n}{S_n^2} \leqslant \frac{a_n}{S_nS_{n-1}}= \frac{S_n - S_{n-1}}{S_nS_{n-1}}= \frac{1}{S_{n-1}}- \frac{1}{S_n}$$
and the RHS has a telescoping sum.
You also need to make it clear that $1/S_n \to 0$ as $n \to \infty$ to argue that the telescoping sum converges to $1/S_1$.
Without nonnegativity it is another story.
