<s>[INST] Prove $\frac{a_n}{S_n^2} \leq \frac{1}{S_{n-1}}-\frac{1}{S_n}$ for partials sums of a divergent series Let $(a_n)$ be a sequence of non-negative numbers such that $a_1 > 0$ and $\sum a_n$ diverges. Let $S_n = \sum_{k=1}^n a_k$.  Prove that, for all $n \geq 2$,
$$\frac{a_n}{S_n^2} \leq \frac{1}{S_{n-1}}-\frac{1}{S_n}$$
How would I start this proof? I've just been staring at it and am very stuck. All i know so far is that $S_n-S_{n-1}=a_n$. Where does the inequality come from?
 [/INST] You can prove this result by induction. First, in the case $n=2$, 
\begin{align*}
\frac{a_2}{S^2_{2}} \leq \frac{1}{S_{1}}-\frac{1}{S_{2}},
\end{align*}
which follows that $\frac{1}{S_{1}}-\frac{1}{S_{2}}= \frac{S_2-S_1}{S_1S_{2}}=\frac{a_2}{S_1S_{2}}$ and $S_2 \geq S_1$.
Suppose that \begin{align*}
\frac{a_n}{S^2_{n}} \leq \frac{1}{S_{n-1}}-\frac{1}{S_{n}}.
\end{align*} 
Then 
\begin{align*}
\frac{a_{n+1}}{S^2_{n+1}} = \frac{a_{n+1}}{(S_{n}+a_{n+1})^2} \leq  \frac{a_{n+1}}{S_{n}S_{n+1}}=\frac{1}{S_{n}}-\frac{1}{S_{n+1}}.
\end{align*}
The proof is completed.
</s>