Does $\sum_k \frac{a_k-a_{k-1}}{a_k^2}$ converge if $a_k \uparrow \infty$?

Question

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?

Answer 1

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

Answer 2

$$ \begin{align} \sum_{k=1}^\infty\frac{a_k-a_{k-1}}{a_k^2} &\le\sum_{k=1}^\infty\frac{a_k-a_{k-1}}{a_ka_{k-1}}\\ &=\sum_{k=1}^\infty\left(\frac1{a_{k-1}}-\frac1{a_k}\right)\\ &=\frac1{a_0}-\lim_{n\to\infty}\frac1{a_n} \end{align} $$