# 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 $$n \le 2$$

$$s_{n-1} \le s_{n}$$ $$\Leftrightarrow \frac{1}{s_{n}} \le \frac{1}{s_{n-1}}$$ $$\Leftrightarrow \frac{1}{s_{n^2}} \le \frac{1}{s_{n}s_{n-1}}$$ $$\Leftrightarrow \frac{a_{n}}{s_{n^2}} \le \frac{a_{n}}{s_{n}s_{n-1}} = \frac{s_{n} - s_{n-1}}{s_{n}s_{n-1}}$$ $$\Leftrightarrow \frac{a_{n}}{s_{n^2}} \le \frac{1}{s_{n-1}} - \frac{1}{s_{n}}$$

• It is a little unclear what your assumptions actually are. – David Peterson Oct 19 '18 at 4:21

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.

• if $a_{n} > 0$ – VERA Oct 19 '18 at 5:11
• @VERA -- yes all $a_n$ positive makes this work. – RRL Oct 19 '18 at 5:12
• You could correct the writing of my proof, please. Because I'm still not very good at writing. – VERA Oct 19 '18 at 5:14
• and make the correction to $\sum_{n=2}^k(1/S_{n-1} - 1/S_n) \neq 1/S_1$ which is $\sum_{n=2}^\infty(1/S_{n-1} - 1/S_n) = 1/S_1$. – RRL Oct 19 '18 at 5:14
• Perfect -- except don't write $1/S_1 - 1/S_k = 1/S_1$ in the third to last sentence. Then its done. – RRL Oct 19 '18 at 5:29