# Infinite Series $\sum\limits_{n=2}^{\infty}\frac{(-1)^n}{n^k}\zeta(n)$

We can prove that $$\sum_{n=2}^{\infty}\frac{(-1)^n}{n}\zeta(n)=\gamma$$

In fact, If we let $f(z)=\sum_{m=2}^\infty\frac{(-1)^m}m z^m$, then by the method which used in this question,

$$\sum_{n=2}^{\infty}\frac{(-1)^n}{n}\zeta(n)=\sum_{n=1}^nf\left(\frac1 n\right)=\sum_{n=1}^\infty\sum_{m=2}^\infty\frac{(-1)^m}{mn^m}=\sum_{n=1}^\infty\left(\frac1 n+\log\left(1-\frac1 n\right)\right)=\gamma$$

Is there any known value for $\displaystyle \sum_{n=2}^{\infty}\frac{(-1)^n}{n^k}\zeta(n)$ for every natural number $k\ge2$? What is the best result?

## migrated from mathoverflow.netOct 19 '13 at 13:25

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I don't know if this can help, probably not, but from $$\sum_{n=2}^{\infty}\frac{\left(-1\right)^{n}x^{n}}{n}\zeta\left(n\right)=x\gamma+\log\left(\Gamma\left(x+1\right)\right)$$ we can get $$\sum_{n=2}^{\infty}\frac{\left(-1\right)^{n}x^{n-1}}{n}\zeta\left(n\right)=\gamma+\frac{\log\left(\Gamma\left(x+1\right)\right)}{x}$$ and so $$\sum_{n=2}^{\infty}\frac{\left(-1\right)^{n}x^{n}}{n^{2}}\zeta\left(n\right)=\gamma x+\int_{0}^{x}\frac{\log\left(\Gamma\left(y+1\right)\right)}{y}dy$$ note that for $x=1$ we have the GEdgar's result. With the same method, we can get $$\sum_{n=2}^{\infty}\frac{\left(-1\right)^{n}x^{n}}{n^{3}}\zeta\left(n\right)=\gamma x+\int_{0}^{x}\frac{1}{t}\left(\int_{0}^{t}\frac{\log\left(\Gamma\left(y+1\right)\right)}{y}dy\right)dt$$ and so on. Surely these integrals are quite frightful.
I only get $$\sum _{n=2}^{\infty }{\frac { \left( -1 \right) ^{n}\zeta \left( n \right) }{{n}^{2}}} = \gamma+\int _{0}^{1}\!{\frac {\ln \left( \Gamma \left( s+1 \right) \right) }{s}}{ds}$$ Probably not much help.
• @draks... take the logarithm of Weierstrass' form of the Gamma function to obtain $\sum_{n=2}^{\infty}\frac{(-1)^n\zeta(n)}{n}x^n=\gamma x+\log\Gamma(x+1)$ – nospoon Sep 22 '15 at 11:00
• @draks... No. divide the above equation by $x$ and integrate from $0$ to $1$. – nospoon Sep 22 '15 at 11:40