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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.

Bernard
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Tom L
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1 Answers1

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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.

Olivier Oloa
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    According to Mathematica: $$\zeta '(2)=\frac{1}{6} \pi ^2 (-12 \log (A)+\gamma +\log (2)+\log (\pi )),$$ where $A$ is Glaisher's Constant and $\gamma$ is the Euler–Mascheroni constant. – James Arathoon Aug 24 '19 at 22:03