Polygamma function series: $\sum_{k=1}^{\infty }\left(\Psi^{(1)}(k)\right)^2$ 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.

Continuing from Olivier Oloa's answer,
$$ \begin{align}  \sum_{k=1}^{\infty} \big(\psi^{(1)} (k) \big)^{2} &= \int_{0}^{1} \int_{0}^{1} \frac{\ln u \ln v}{(1-uv)(1-u)(1-v)} \ du \ dv \\ &= \int_{0}^{1} \frac{\ln v}{(1-v)^{2}} \int_{0}^{1} \left(\frac{\ln u}{1-u} - \frac{v \ln u}{1-vu} \right) \ du \ dv \\ &= \int_{0}^{1} \frac{\ln v}{(1-v)^{2}} \left( \int_{0}^{1} \frac{\ln u}{1-u} \ du - v \int_{0}^{1} \frac{\ln u}{1-vu} \ du\right) \ dv \end{align}$$
where $$  \int_{0}^{1} \frac{\ln u}{1-u} \ du = \int_{0}^{1} \frac{\ln (1-w)}{w} = -\text{Li}_{2}(1) = -\zeta(2) $$
and
$$  \int_{0}^{1} \frac{\ln u}{1-vu} \ du = - \frac{1}{v}  \ln(1-vu) \ln u \Bigg|^{1}_{0} + \frac{1}{v} \int_{0}^{1} \frac{\ln (1-vu)}{u} \ du  = - \frac{\text{Li}_{2}(v)}{v} .$$
Therefore,
$$ \sum_{k=1}^{\infty} \big(\psi^{(1)} (k) \big)^{2} = \int_{0}^{1} \frac{\ln v}{(1-v)^2} \Big(\text{Li}_{2}(v) - \zeta(2) \Big) \ dv .$$
Then integrating by parts,
$$ \begin{align} &= \big(\text{Li}_{2}(v) - \zeta(2) \big) \left(\ln (1-v) + \frac{v \ln v}{1-v} \right)\Bigg|^{1}_{0} + \int_{0}^{1} \frac{\ln^{2}(1-v)}{v} \ dv + \int_{0}^{1} \frac{\ln(1-v) \ln v}{1-v} \ dv \\ &=  \int_{0}^{1} \frac{\ln^{2}(1-v)}{v} \ dv + \int_{0}^{1} \frac{ \ln(1-v) \ln v}{1-v} \ dv \end{align}$$
where $$ \begin{align} \int_{0}^{1} \frac{\ln^{2}(1-v)}{v} \ dt &= \ln^{2}(1-v)\ln v \Bigg|^{1}_{0} + 2 \int_{0}^{1} \frac{\ln(1-v) \ln v}{1-v} \ dv \\ &=  2 \int_{0}^{1} \frac{ \ln(1-v) \ln v}{1-v} \ dv . \end{align} $$
Therefore, $$ \begin{align} \sum_{k=1}^{\infty} \big(\psi^{(1)} (k) \big)^{2} &= 3 \int_{0}^{1} \frac{\ln(1-v) \ln v}{1-v} \ dv \\ &= -3 \int_{0}^{1} \ln v \sum_{n=1}^{\infty} H_{n}v^{n} \ dv  \\ & = -3 \sum_{n=1}^{\infty} H_{n} \int_{0}^{1} v^{n} \ln v \ dv \\ &= 3 \sum_{n=1}^{\infty} \frac{H_{n}}{(n+1)^{2}} \\ &= 3 \left(\sum_{n=1}^{\infty} \frac{H_{n+1}}{(n+1)^{2}}  - \sum_{n=1}^{\infty} \frac{1}{(n+1)^{3}} \right) \\ &= 3 \left( \sum_{n=1}^{\infty} \frac{H_{n}}{n^{2}} -1  - \zeta(3) + 1 \right) \\ &= 3 \big(2 \zeta(3) - \zeta(3) \big) \tag{1} \\ &= 3 \zeta(3)  .\end{align} $$
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$(1)$ Generalized Euler sum $\sum_{n=1}^\infty \frac{H_n}{n^q}$
