I want to know if there exists, and how to arrive at, a closed form of this infinite sum: $$S_a=\sum_{n=1}^\infty \frac{\ln(n)}{n^a}$$ I know the series converges at least for every $a>1$ by the comparison test since $$S_a\le\int_1^\infty \frac{\ln(x)}{x^a}dx=\frac{1}{\left(1-a\right)^2}$$ But I have been thinking hard and I'm not reaching any results. The furthest I've gone is saying:$$\sum_{n=1}^{\infty}\frac{1}{n^a} \ln\left(\frac{e}{n}\right)=\zeta(a)-S_a$$ Does anyone have any hints?
Thank you!

  • $\begingroup$ It's essentially the derivative of the zeta function. $\endgroup$ – Marco Cantarini Nov 20 '15 at 19:44

One may recall that the Riemann zeta function $$ \zeta(a)=\sum_{n=1}^\infty\frac1{n^a} $$ is differentiable on $(1,\infty)$ admitting a derivative given by $$ \zeta'(a)=-\sum_{n=1}^\infty\frac{\ln n}{n^a}. $$


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