# On the series $\sum \limits_{n=2}^{\infty} \frac{(-1)^n}{n \log n}$

The series $\displaystyle \sum_{n=2}^{\infty} \frac{1}{n \log n}$ is known to diverge. This can be seen with comparison to the harmonic series.

The other series $\displaystyle \sum_{n=2}^{\infty} \frac{(-1)^n}{n \log n}$ converges due to Liebniz's alternating criterion. I see these two series quite often in exams' questions to examine them if they converge or not. The answers are pretty straight forward.

Now , out of curiosity , could we find a closed form for the series

$$\mathcal{S}=\sum_{n=2}^{\infty} \frac{(-1)^n}{n \log n}$$

The interesting fact is that if we define the function

$$f(s)=\sum_{n=2}^{\infty} \frac{(-1)^n}{n^s \log n} , \; s \geq 1$$

then taking the derivative with respect to $s$ we get that

$$f'(s)=-\sum_{n=2}^{\infty} \frac{(-1)^n}{n^s} = \zeta(s) - 2^{1-s} \zeta(s) - 1$$

I don't know how to go back and evaluate $f(1)$. Has anyone done that before. Expressing the value of the series in terms of special functions will suit me. Numerically it is approximately equal to $0.526412$. That said Wolfram. So, any clues?

• I don't think there's a known closed form expression to $f(1)$. – vrugtehagel Mar 9 '17 at 9:50
• I was afraid of that. I did not actually expect to be any but that was just me. Some people here are amazing finding the craziest closed expressions to such problems. – Tolaso Mar 9 '17 at 9:58
• Well you may deduce that $\;f(1)=\displaystyle\int_1^\infty 1+\left(2^{1-s}-1\right)\zeta(s)\,ds\;$ but I wouldn't bet on a closed form... – Raymond Manzoni Mar 9 '17 at 10:15
• Yeah I had considered that as well but gave up since I don't know how to integrate zeta function. Oops.!! – Tolaso Mar 9 '17 at 10:27
• I would guess this is is as difficult as math.stackexchange.com/questions/741526/… since $H_n \sim \log (n)$ near infinity. – Zaid Alyafeai Aug 6 '17 at 17:48