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?
 A: Some manipulation of the inequality
$$\frac{a_n}{S_n^2} \leq \frac{1}{S_{n-1}} - \frac{1}{S_n} = \frac{S_n - S_{n - 1}}{S_{n-1}S_n} = \frac{a_n}{S_{n-1}S_n}$$
Since $a_n \geq 0$ this is equivalent to
$$S_{n-1}S_n \leq S_n^2$$
Since $S_n \gt 0$ this is equivalent to
$$S_{n-1} \leq S_n$$
This is true since $S_n - S_{n-1} = a_n \geq 0$.
A: A start: Note that 
$$a_n=S_n-S_{n-1}.\tag{1}$$
That holds because 
$$S_n=(a_1+\cdots+a_{n-1})+a_n=S_{n-1}+a_n.$$
The equality (1) and some manipulation is all you will need. Bring your right-hand side to a common denominator.
A: Since $a_n$ is nonnegative, $S_n$ is non-decreasing.
Thus $S_{n-1} \leqslant S_n \implies 1/S_n \leqslant 1/S_{n-1},$ and
$$\frac{a_n}{S_n^2} = \frac{S_n - S_{n-1}}{S_n^2}\leqslant \frac{S_n - S_{n-1}}{S_{n-1}S_n} = \frac1{S_{n-1}} - \frac1{S_n}.$$
Hence, with $a_1 > 0$ we have
$$\begin{align}\sum_{n=1}^m \frac{a_n}{S_n^2} &= \frac1{a_1} + \sum_{n=2}^m \frac{a_n}{S_n^2} \\ &\leqslant  \frac1{a_1} + \sum_{n=2}^m  \left(\frac1{S_{n-1}} - \frac1{S_n} \right) \\ &= \frac1{a_1} + \frac{1}{S_1} - \frac{1}{S_m} \\ & \leqslant \frac{2}{a_1}.\end{align}$$
Thus, the sequence of partial sums is bounded and increasing (since the terms are non-negative) and the series converges.
A: 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.
A: You have $S_n^2(1/S_n-1/S_{n-1})\ge a_n$, or $S_n/S_{n-1}(S_n-S_{n-1}) \ge a_n$. Ultimately, $S_n/S_{n-1}a_n \ge a_n$ or $S_n \ge S_{n-1}$, which is true, since $S_n-S_{n-1} = a_n > 0$. Hope this helps.
A: $$\frac{a_n}{S_n^2} \leq \frac{1}{S_{n-1}}-\frac{1}{S_n} = \frac{S_n - S_{n-1}}{S_{n-1} S_n} = \frac{a_n}{S_{n-1} S_n}$$
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
$$0 \leq a_n$$
