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.
 A: $$\begin{align*}
\sum_{n=1}^{\infty} \frac{1}{n^2}\left(\sum_{i=1}^n \frac{1}{\sqrt{i}}\right) &= \sum_{i=1}^{\infty} \frac{1}{\sqrt{i}}\left(\sum_{n=i}^{\infty} \frac{1}{n^2}\right) \\
&< \frac{\pi^2}{6} + \sum_{i=2}^{\infty}\frac{1}{\sqrt{i}}\left(\sum_{n=i}^{\infty} \frac{1}{n(n-1)}\right) \\
&=\frac{\pi^2}{6} + \sum_{i=2}^{\infty}\frac{1}{\sqrt{i}}\left(\sum_{n=i}^{\infty} \frac{1}{n - 1} - \frac{1}{n}\right) \\
&=\frac{\pi^2}{6} + \sum_{i=2}^{\infty}\frac{1}{\sqrt{i}}\cdot\frac{1}{i - 1} \\
&<\frac{\pi^2}{6} + \sum_{i=2}^{\infty}\frac{1}{\sqrt{i}}\cdot\frac{2}{i}
\end{align*}$$
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.
A: 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.
A: 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 $3.44844$
A: Just for fun.
By the Hermite-Hadamard inequality
$$ \sqrt{4n+2}-\sqrt{6}=\int_{3/2}^{n+1/2}\frac{dx}{\sqrt{x}}\geq \sum_{k=2}^{n}\frac{1}{\sqrt{k}} $$
and by the inequalities fulfilled by the central binomial coefficients
$$ \sqrt{4n+2}\leq \left(2+\frac{1}{4n}\right)\frac{4^n}{\sqrt{\pi}\binom{2n}{n}} $$
so the following approximation holds:
$$ \sum_{n=1}^{+\infty}\frac{1}{n^2}\sum_{k=1}^{n}\frac{1}{\sqrt{k}} \leq \sqrt{6}-\frac{\pi^2}{6}(\sqrt{6}-1)+\pi^{3/2}+\frac{\pi^{3/2}}{4}\log(2)-\frac{36+7\zeta(3)}{8\sqrt{\pi}}<\color{red}{\frac{52}{15}} .$$
