Convergence of $\sum_{n=1}^\infty \frac{\sum_{i=1}^n\frac{1}{ \sqrt i}}{n^2}$ How can I prove the following sequence converges?
$$\sum_{n=1}^\infty \frac{\sum_{i=1}^n\frac{1}{ \sqrt i}}{n^2}$$
I tried everything.
Could not find any candidates for comparison test, and failed to find an upper bound.
Any hints will be appreciated.

By Cauchy-Schwarz inequality:$$\begin{align}a_n&= \dfrac{1+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+\cdots + \dfrac{1}{\sqrt{n}}}{n^2}\\&\le \dfrac{\sqrt{n}\sqrt{1+\dfrac{1}{2}+\dfrac{1}{3}+\cdots +\dfrac{1}{n}}}{n^2}\\&\le \dfrac{\sqrt{\ln(n+1)}}{n^{\frac{3}{2}}}\\&\le \dfrac{(n+1)^c}{n^{\frac{3}{2}}}\\&= \left(1+\dfrac{1}{n}\right)^c\cdot \dfrac{1}{n^{\frac{3}{2}-c}}\\&\le \dfrac{2^c}{n^{\frac{3}{2} - c}}\\&=b_n.\end{align}$$You can choose the constant $c > 0$ such that $\dfrac{3}{2} - c > 1$ or $0 < c < \dfrac{1}{2}$. Thus $\displaystyle \sum_{n=1}^\infty b_n$ converges by the  comparing it to the $p$-series for $p > 1$. And by comparison test once again to the $b_n$-series, $\displaystyle \sum_{n=1}^\infty a_n$ converges.
