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}} $

  • $\begingroup$ It is a little unclear what your assumptions actually are. $\endgroup$ – 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.

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

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