Series involving Riemann zeta function

[Grigorios Kostakos]

Prove that \[\displaystyle\sum_{n=2}^\infty \frac{(-1)^n\,\zeta(n)}{n\,(n+1)}=\frac{1}{2}(\log2+\log\pi+\gamma-2)\] where \(\zeta\) is Riemann zeta function and \(\gamma\) is Euler–Mascheroni constant.



P.S. Mentioned by akotronis on Integral involving $\Gamma$ function.

[akotronis]

We have \(\begin{align*}\displaystyle\sum_{n\geq2}\frac{(-1)^n\zeta(n)}{n(n+1)}=\sum_{n\geq2}\frac{(-1)^n\zeta(n)}{n}-\sum_{n\geq2}\frac{(-1)^n\zeta(n)}{n+1}:=A-B\end{align*}\).

Now from the Taylor series of \(\ln(1-x)\) it is \(\begin{align}\label{1}\displaystyle\sum_{n\geq2}\frac{x^n}{n}&=-\ln(1-x)-x,\hspace{10ex}&x\in[-1,1)\tag{1}\qquad\text{and}\\\label{2}\sum_{n\geq2}\frac{x^n}{n+1}&=-\frac{\ln(1-x)}{x}-1-\frac{x}{2},\hspace{10ex}&x\in[-1,1)\tag{2}.\end{align}\)

For B:

\(\begin{align*}B&=\sum_{n\geq2}\frac{(-1)^n}{n+1}\left(1+\sum_{k\geq1}\frac{1}{k^n}\right)\\&\stackrel{*}{=}\sum_{n\geq2}\frac{(-1)^n}{n+1}+\sum_{k\geq2}\sum_{n\geq2}\frac{(-1/k)^n}{n+1}\\&\stackrel{(2)}{=}\ln2-\frac{1}{2}+\sum_{k\geq2}\frac{1}{2k}-1+k\ln\left(1+\frac{1}{k}\right)\\&=\ln2-\frac{1}{2}+\lim_{n\to+\infty}\left(\frac{1}{2}(H_n-1)-(n-1)+\ln\prod_{k=2}^{n}\frac{(k+1)^{k}}{k^k}\right)\\&=\ln2-\frac{1}{2}+\lim_{n\to+\infty}\left(\frac{1}{2}(H_n-1)-n+1+\ln\prod_{k=2}^{n}\frac{(k+1)^{k+1}}{k^k}\frac{1}{k+1}\right)\\&=\ln2-\frac{1}{2}+\lim_{n\to+\infty}\left(\frac{1}{2}(H_n-1)-n+1+\ln\frac{2}{(n+1)!}+\ln\frac{(n+1)^{n+1}}{4}\right)\\&\to1+\frac{\gamma}{2}-\frac{\ln2}{2}-\frac{\ln\pi}{2},\end{align*}\)

where we used stirling's approximation and that \(H_n=\ln n+\gamma+\mathcal O(n^{-1})\). (* the change of the summation order is justified due to absolute convergence).

With exactly the same method, and using (1) this time, we find that \(A=\gamma\) and the result follows.

A similar problem which is solved with the same technique can be found in the current issue of MATHPROBLEMS magazine here (http://mathproblems-ks.com/?wpfb_dl=8) page 121.

For generalizations of this problem and the connection it has with the topic discussed here (Integral involving $\Gamma$ function), one can see at the articles

1) J.W.L. Glaisher, Relations connecting quantities of the form \(1 + 2^{-n} + 3^{-n} + 4^{-n}+\cdots\) Messenger of Mathematics 44 (1914 - 1915) 1 - 10. (http://archive.org/details/messengerofmathe44cambuoft) and

2) S. Ramanujan, A series for Euler’s constant, Messenger of Mathematics 46 (1917) 73 - 80. (http://archive.org/details/messengerofmathe46cambuoft)

[Tolaso J Kos]

Prove that \[\displaystyle\sum_{n=2}^\infty \frac{(-1)^n\,\zeta(n)}{n\,(n+1)}=\frac{1}{2}(\log2+\log\pi+\gamma-2)\] where \(\zeta\) is Riemann zeta function and \(\gamma\) is Euler–Mascheroni constant.



P.S. Mentioned by akotronis on Integral involving $\Gamma$ function.

Here is another solution.

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 $$\sum_{n=2}^\infty\frac{(-1)^n \zeta(n)}{n(n+1)}=\int_{0}^{1}(\gamma x+\log(x)+\log\Gamma(x))\,{\rm d}x=\frac{\gamma}{2}-1+\int_{0}^{1}\log\Gamma(x)\,{\rm d}x$$

$\int_{0}^{1}\log\Gamma(x)dx=\log\sqrt{2\pi}$ has been proved above at the topic linked at the first post.