TITLE: $ \sum_{k=1}^{\infty}\frac{\log(k)}{k^2(k+1)}$ explicit value.
QUESTION [5 upvotes]: I am trying to evaluate the following series explicitly, 
$$ \sum_{k=1}^{\infty}\frac{\log(k)}{k^2(k+1)}$$
I know this converges by the comparison test.
I have tried defining a function,
$$f(t)=\sum_{k=1}^{\infty}\frac{\log(kt)}{k^2(k+1)}$$
Differentiating this with respect to t gives,
$$f'(t)=\frac{1}{t}\sum_{k=1}^{\infty} \frac{1}{k^2(k+1)} = \frac{1}{t}(\zeta(2)-1)$$
Integrating this,
$$\int_a^t f'(s)ds = f(t)-f(a) = (\zeta(2)-1)(\log(t) - \log(a))$$
for some constant a>0. However whichever constant I choose I get in a bit of a muddle when trying to evaluate at t=1. I was hoping for a hint in the right direction or maybe a suggestion of a different method that might be useful.

REPLY [5 votes]: The given series admits a closed form in terms of the poly-Stieltjes constants defined here. By using the absolute convergence, on may write 
\begin{align}
\sum_{k=1}^{\infty}\frac{\log(k)}{k^2(k+1)}&=\sum_{k=1}^{\infty}\frac{\log(k)}{k^2}-\sum_{k=1}^{\infty}\frac{\log(k)}{k(k+1)}
=-\zeta'(2)-\gamma_1+\gamma_1(0,1),
\end{align} having applied Theorem 2 to the latter series.