TITLE: Polygamma function series: $\sum_{k=1}^{\infty }\left(\Psi^{(1)}(k)\right)^2$
QUESTION [29 upvotes]: Applying the Copson's inequality, I found:
$$S=\displaystyle\sum_{k=1}^{\infty }\left(\Psi^{(1)}(k)\right)^2\lt\dfrac{2}{3}\pi^2$$ where
$\Psi^{(1)}(k)$ is the polygamma function.
Is it known any sharper bound for the sum $S$?
Thanks.

REPLY [33 votes]: First, we have
$$
\begin{align}
\psi'(n)
&=\sum_{k=0}^\infty\frac1{(k+n)^2}\\
&=\sum_{k=n}^\infty\frac1{k^2}\tag{1}
\end{align}
$$
Then
$$
\begin{align}
\sum_{n=1}^\infty\psi'(n)^2
&=\sum_{n=1}^\infty\sum_{j=n}^\infty\frac1{j^2}\sum_{k=n}^\infty\frac1{k^2}\tag{2}\\
&=\sum_{n=1}^\infty\left(\sum_{j=n}^\infty\frac1{j^4}+2\sum_{j=n}^\infty\sum_{m=1}^\infty\frac1{j^2}\frac1{(j+m)^2}\right)\tag{3}\\
&=\sum_{j=1}^\infty\sum_{n=1}^j\frac1{j^4}+2\sum_{j=1}^\infty\sum_{m=1}^\infty\sum_{n=1}^j\frac1{j^2}\frac1{(j+m)^2}\tag{4}\\
&=\sum_{j=1}^\infty\frac1{j^3}+2\sum_{j=1}^\infty\sum_{m=1}^\infty\frac1{j(j+m)^2}\tag{5}\\
&=\zeta(3)+2\sum_{n=1}^\infty\frac{H_{n-1}}{n^2}\tag{6}\\
&=\zeta(3)-2\zeta(3)+2\sum_{n=1}^\infty\frac{H_n}{n^2}\tag{7}\\[9pt]
&=\zeta(3)-2\zeta(3)+4\zeta(3)\tag{8}\\[18pt]
&=3\zeta(3)\tag{9}
\end{align}
$$
Explanation:
$(2)$: use $(1)$
$(3)$: first sum covers $j=k$ the other $j\lt k$ and $j\gt k$
$(4)$: change order of summation
$(5)$: sum in $n$
$(6)$: $n=j+m$ and $j=n-m$ and sum in $m$
$(7)$: $H_{n-1}=H_n-\frac1n$
$(8)$: equation $(14)$ of this answer with $q=2$
$(9)$: add