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.

Using generalized harmonic numbers, you face
$$S_p=\sum_{n=1}^p \frac{H_n^{\left(\frac{1}{2}\right)}}{n^2}$$
For large values of $n$
$$H_n^{\left(\frac{1}{2}\right)}=2\sqrt{n}+\zeta \left(\frac{1}{2}\right)+\frac{1}{2 \sqrt{n}}\Bigg[1-\frac{1}{12 n}+\frac{1}{192 n^3}-\frac{1}{512
   n^5}+O\left(\frac{1}{n^6}\right) \Bigg]$$ So, the summation does converge.
Using only the above terms
$$\sum_{n=1}^\infty \frac{H_n^{\left(\frac{1}{2}\right)}}{n^2}=\frac{1}{6} \pi ^2 \zeta \left(\frac{1}{2}\right)+2 \zeta
   \left(\frac{3}{2}\right)+\frac{\zeta \left(\frac{5}{2}\right)}{2}-\frac{\zeta
   \left(\frac{7}{2}\right)}{24}+\frac{\zeta
   \left(\frac{11}{2}\right)}{384}-\frac{\zeta \left(\frac{15}{2}\right)}{1024}$$
which is $3.44805$ while numerically, the infinite summation is $3.44844$
