TITLE: Prove $\frac{a_n}{S_n^2} \leq \frac{1}{S_{n-1}}-\frac{1}{S_n}$ for partials sums of a divergent series
QUESTION [7 upvotes]: 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?

REPLY [2 votes]: $$\frac{a_n}{S_n^2} \leq \frac{1}{S_{n-1}}-\frac{1}{S_n} = \frac{S_n - S_{n-1}}{S_{n-1} S_n} = \frac{a_n}{S_{n-1} S_n}$$
Multiply both sides by $q=S_n/a_n$:
$$\frac 1{S_n} \leq \frac 1{S_{n-1}}$$
For given properties of $(a_n)$ the value of $q$ is postive, so the inequality does not change the direction. For the same reason we can multiply now by both denominators:
$$S_{n-1} \leq S_n$$
and after subtraction
$$0 \leq S_n - S_{n-1}$$
which is
$$0 \leq a_n$$