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Let $a_n$ be a sequence of strictly increasing postive numbers such that $a_n \uparrow \infty$. Does $$\sum_{k=1}^\infty \frac{a_k-a_{k-1}}{a_k^2}$$ converge?

My guess is that it converges. I tried with $a_n=n, n^2, \log n$. For all of them the series converges.

My try:

I was thinking of using Cauchy–Schwarz inequality, like $$\sum_{k=1}^\infty \frac{a_k-a_{k-1}}{a_k^2} \le \sum_{k=1}^\infty \frac1{a_k^2}\sum_{k=1}^\infty \frac{(a_k-a_{k-1})^2}{a_k^2}$$ or some other variants. But none of them works.

Then I tried to show that the series is Cauchy, i.e., for $p<q$ $$\sum_{k=p+1}^q \frac{a_k-a_{k-1}}{a_k^2} \le \sum_{k=p+1}^q \frac{a_k-a_{k-1}}{a_{p+1}^2}=\frac{a_q-a_p}{a_{p+1}^2}$$ but the right hand estimate does not necessarily go to zero for large $p$ and $q$. Take $a_n=n^n$ for example. But the original series converges with $a_n=n^n$.

Any help/suggestions?

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Note $a_k^2 > a_k a_{k-1}$.

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  • $\begingroup$ I feel so stupid! $\endgroup$ – Sayan May 31 '17 at 1:28
  • $\begingroup$ chill sayan :p I feel stupid all the time :p $\endgroup$ – Arpan1729 May 31 '17 at 15:37
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$$ \begin{align} \sum_{k=1}^\infty\frac{a_k-a_{k-1}}{a_k^2} &\le\sum_{k=1}^\infty\frac{a_k-a_{k-1}}{a_ka_{k-1}}\\ &=\sum_{k=1}^\infty\left(\frac1{a_{k-1}}-\frac1{a_k}\right)\\ &=\frac1{a_0}-\lim_{n\to\infty}\frac1{a_n} \end{align} $$

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  • $\begingroup$ @MarkViola: Arin Chaudhuri's hint gives the essence, and it was first. $\endgroup$ – robjohn May 31 '17 at 3:00
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    $\begingroup$ That's fine. From the time stamp, they appear to have been contemporaneous. $\endgroup$ – Mark Viola May 31 '17 at 3:04

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