Suppose $a_n>0$ and $\sum_{n=1}^{\infty}{a_n}$ diverges. Determine convergence of $\sum_{n=1}^{\infty}{\frac{a_n}{s_n^2}}$, where $s_n=\sum^n a_n$.

Question

Suppose $a_n>0$ and $\sum_{n=1}^{\infty}{a_n}$ diverges. Determine whether $\sum_{n=1}^{\infty}{\frac{a_n}{s_n^2}}$ converges, where $s_n=a_1+a_2+ \cdots + a_n$.

My attempt:

By testing a few examples, the series $\sum_{n=1}^{\infty}{\frac{a_n}{s_n^2}}$ converges. We proceed to prove it.

Note that

$$\frac{a_n}{s_n^2} \leq \frac{a_n}{n^2(a_1a_2\cdots a_n)}$$

Now if I manage to prove that $a_1a_2\cdots a_n \geq 1$, then the inequality above becomes

$$\frac{a_n}{s_n^2} \leq \frac{1}{n^2}$$. My guess is that it should have something to do with the divergent series $\sum_{n=1}^{\infty}{a_n}$.

Then by the Comparison test, we are done. However, I have difficulty to prove the claim. Can anyone give some hint?

UPDATE: So I made some mistake in my working. Here is my another 'promising' claim: $$a_n \leq (\frac{a_1+...+a_n}{n})^2$$ It seems to work for any series satisfying the question. But I am unable o prove it.

Answer 1

\begin{align} \sum_{n=p+1}^{m}{\frac{a_n}{s_n^2}}&=\sum_{n=p+1}^{m}{\frac{s_n-s_{n-1}}{s_n^2}} \\ &=\sum_{n=p+1}^{m}\frac{s_n-s_{n-1}}{s_{n-1}s_n}\frac{s_{n-1}}{s_n} \\ &=\sum_{n=p+1}^{m}\frac{s_{n-1}}{s_n}\left(\frac{1}{s_{n-1}}-\frac{1}{s_n}\right) \\ &< \sum_{n=p+1}^{m}\left(\frac{1}{s_{n-1}}-\frac{1}{s_n}\right) \hspace{8 mm} \left(0<\frac{s_{n-1}}{s_n}<1\operatorname{ and }\frac{1}{s_{n-1}}-\frac{1}{s_n}>0\right) \\ &=\frac{1}{s_{p}}-\frac{1}{s_m} \\ &<\frac{1}{s_{p}}\to0 \hspace{8 mm} \left(s_{p} \to\infty \operatorname{ as } p\to\infty\right) \end{align} So by Cauchy Criterion, $\sum \limits_{n=1}^{\infty}{\dfrac{a_n}{s_n^2}}$ converges.

Answer 2

Here's an idea: Suppose $f$ is positive and continuous on $[1,\infty).$ The integral analogue of our problem is: If $\int_1^\infty f = \infty,$ and $F(x) = \int_1^x f,$ then

$$\int_2^\infty \frac{f(x)}{(F(x))^2}\,dx < \infty.$$

This is simple to verify, since $f= F'.$ That strongly suggests $\sum (a_n/s_n^2) < \infty$ in the series case.

Answer 3

For $n \ge 2$ you have, since $a_n>0 $ and $(s_n) $ is non negative and increasing : $$\frac{a_n}{s_n^2} = \frac{s_n-s_{n-1}}{ s_n^2}\le \frac{s_n-s_{n-1}}{ s_{n-1}^2} \le \int_{s_{n-1}}^{s_n} \frac{dt}{t^2} $$ Therefore, for all $N \ge 2$ :

$$ \sum_{n=2}^N \frac{a_n}{s_n^2} \le \int_{s_1}^{s_N} \frac{dt}{t^2} = \left[ -\frac{1}{t} \right]_{s_1}^{s_N} = \frac{1}{s_1} - \frac{1}{s_N} \le \frac{1}{s_1} $$

The partial sum is bounded and the terms are non negetive so the serie $\sum \frac{a_n}{s_n^2} $ is convergent.

Answer 4

The series $\sum \frac{a_n}{s_n^2}$ converges provided each $a_n>0$; no other assumption is needed. Set aside the $n=1$ term, which equals $\frac1{a_1}$. For $n\ge2$, observe: $$ 0<\frac{a_n}{s_n^2}=\frac{s_n-s_{n-1}}{s_ns_n}\le\frac{s_n-s_{n-1}}{s_{n-1}s_n}=\frac1{s_{n-1}}-\frac1{s_n} $$ and by telescoping deduce $$ \sum_{n=2}^N\frac{a_n}{s_n^2}\le\frac1{s_1}-\frac1{s_N}\le\frac1{s_1}=\frac1{a_1}. $$ So the partial sums of the series are bounded above by $\frac2{a_1}$. Since the terms are positive, the partial sums are increasing. Conclude that the series converges.

Answer 5

Let's show that for ${t}>1$ the series $$\sum_n \frac{a_n}{s_n^{{t}}}$$ is convergent.

Apply Lagrange intermediate value theorem for the function $x \mapsto x^{1-{t}}$ on the interval $[s_{n-1}, s_n]$ and we get $$s_{n-1}^{1-{t}}- s_n^{1-{t}} = \frac{({t}-1)(s_n - s_{n-1})}{\theta^{{t}}} \ge ({t}-1) \frac{a_n}{s_n^{{t}}}$$ and it follows that $$\sum_{n\ge 1} \frac{a_n}{s_n^{{t}}}= a_1^{1-t} + \sum_{n\ge 2} \frac{a_n}{s_n^{{t}}} \le a_1^{1-t}+ \frac{1}{t-1} \cdot \sum_{n\ge 2} (s_{n-1}^{1-{t}}- s_n^{1-{t}}) = \frac{t}{t-1} a_1^{1-t}$$ hence the series has a finite sum.