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Can anyone help me attain the result for the following series?

$$\sum_{n=2}^{\infty} \frac{(-1)^n \zeta(n)}{n(n+1)}= \frac{1}{2} \left( \log 2 + \log \pi +\gamma -2 \right)$$

I don't know how to start. I am seriously thinking that this can be done using residues or contour integration since with real analysis I cannot see a pattern.

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    $\begingroup$ Can this be of help? Not exactly the same, but it might give you some inspiration. $\endgroup$
    – Daniel R
    Jun 5, 2015 at 7:48
  • $\begingroup$ Perhaps, I'll take a look! Thanks! $\endgroup$
    – Tolaso
    Jun 5, 2015 at 8:42
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    $\begingroup$ @DanielR It seems like a good start. $\endgroup$
    – Tolaso
    Jun 5, 2015 at 8:46
  • $\begingroup$ Choi and Srivastava show in dx.doi.org/10.1006/jmaa.1998.6216 that $\sum_{k\ge 2} (-1)^k\frac{\zeta(k)}{k(k+1)}z^{k+1} =[\log(2\pi)-1]\frac{z}{2}+(\gamma-1)\frac{z^2}{2} +z\log\Gamma(1+z)-\log G(1+z)$ $\endgroup$ Jun 30, 2021 at 13:21

2 Answers 2

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Take the logarithm of the Weierstass product form of the Gamma function $$ \frac{1}{\Gamma(x)}=xe^{\gamma x}\prod_{n=1}^\infty \left( 1+\frac{x}{n}\right)e^{-x/n}$$ to obtain $$-\log\Gamma(x)=\log x+\gamma x+\sum_{n=1}^\infty \log\left(1+\frac{x}{n}\right)-\frac{x}{n}$$ Now, $$\sum_{n=1}^\infty \log\left(1+\frac{x}{n}\right)-\frac{x}{n} =-\sum_{n=1}^\infty\sum_{m=2}^\infty\frac{\left(-\frac{x}{n}\right)^m}{m} \\\\=-\sum_{m=2}^\infty\frac{(-1)^m x^m}{m}\sum_{n=1}^\infty\frac{1}{n^m}=-\sum_{m=2}^\infty\frac{(-1)^m x^m \zeta(m)}{m}$$ so that $$\sum_{n=2}^\infty\frac{(-1)^n \zeta(n)}{n}x^n=\gamma x+\log(x)+\log\Gamma(x)$$ and $$\sum_{n=2}^\infty\frac{(-1)^n \zeta(n)}{n(n+1)}=\int_{0}^{1}(\gamma x+\log(x)+\log\Gamma(x))dx=\frac{\gamma}{2}-1+\int_{0}^{1}\log\Gamma(x)dx$$ $\int_{0}^{1}\log\Gamma(x)dx=\log\sqrt{2\pi}$ is quite famous. check here for reduction to evaluating $\int_0^\pi \log\sin(x)dx$ and here for the evaluation

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    $\begingroup$ That is amazing. Thank you. I manages to derive the identity: $$\sum_{n=2}^\infty\frac{(-1)^n \zeta(n)}{n}x^n=\gamma x+\log(x)+\log\Gamma(x)$$ using the link above, but then I was too tired to construct the wanted series. (actuall I did not see that it was related to the integral) $\endgroup$
    – Tolaso
    Jun 5, 2015 at 11:54
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it it not difficult to show that $$\sum _{k=2}^{\infty } \frac{\pi (-x)^k \zeta (k)}{k (k+1)}=\int_0^{\infty } -\frac{(t-2 i \pi x) \log \left(1+\frac{i t}{2 \pi x}\right)+(t+2 i \pi x) \log \left(1-\frac{i t}{2 \pi x}\right)-2 t}{\left(e^t-1\right) (4 \pi x)} \, dt+\frac{1}{2} (-\pi \log (x)+\pi -\pi \log (2 \pi ))+\frac{1}{4} (-2 \gamma \pi x+3 \pi x-2 \pi x \log (x))$$

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