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.

$$\begin{align*}
\sum_{n=1}^{\infty} \frac{1}{n^2}\left(\sum_{i=1}^n \frac{1}{\sqrt{i}}\right) &= \sum_{i=1}^{\infty} \frac{1}{\sqrt{i}}\left(\sum_{n=i}^{\infty} \frac{1}{n^2}\right) \\
&< \frac{\pi^2}{6} + \sum_{i=2}^{\infty}\frac{1}{\sqrt{i}}\left(\sum_{n=i}^{\infty} \frac{1}{n(n-1)}\right) \\
&=\frac{\pi^2}{6} + \sum_{i=2}^{\infty}\frac{1}{\sqrt{i}}\left(\sum_{n=i}^{\infty} \frac{1}{n - 1} - \frac{1}{n}\right) \\
&=\frac{\pi^2}{6} + \sum_{i=2}^{\infty}\frac{1}{\sqrt{i}}\cdot\frac{1}{i - 1} \\
&<\frac{\pi^2}{6} + \sum_{i=2}^{\infty}\frac{1}{\sqrt{i}}\cdot\frac{2}{i}
\end{align*}$$
which converges.
For those interested, the bound I gave above is $\frac{\pi^2}{6} + \zeta\left(\frac{3}{2}\right) - 1 \approx 6.86968476421920$, while the correct sum seems to be about $3.44\ldots$ by summing the first $10^6$ terms. I wonder what the correct sum is.
If you don't understand the first line, try drawing $16$ points in a $4$ by $4$ square, label them $(n, i) = (1, 1)$ to $(n, i) = (4, 4)$ from bottom left to top right, then colour/circle the points that will be included in the sum. You should colour the points $i \leq n$ only.
