Sum with Riemann zeta function

Question

I found on Instagram an interesting series: $$\sum_{n\geq1} \zeta(n+1)\frac{n}{2^n}$$ In a first moment I noticed that it can be written as $$\sum_{n\geq1}\zeta(n+1)\Gamma(n+1)\frac1{2^n}$$ and so to use the identity involving both Zeta and Gamma functions $$\zeta(s)\Gamma(s)=\int_0^\infty \frac{x^{s-1}}{e^x-1}\,\mathrm{d}x$$ The problem in this idea comes with the radius of convergence of $\sum_{n\geq1}(\frac x2)^n$ that isn't $(0,\infty)$ at all… Have you some other approach to this little monster?

Answer 1

For $|x|\lt1$, we have

$$\sum_{n=1}^\infty\zeta(n+1)x^n=\sum_{n=1}^\infty\sum_{k=1}^\infty{x^n\over k^{n+1}}=\sum_{k=1}^\infty\sum_{n=1}^\infty{1\over k}\left(x\over k\right)^n=\sum_{k=1}^\infty{1\over k}\cdot{x/k\over1-x/k}=\sum_{k=1}^\infty{x\over k(k-x)}$$

and since ${x\over k-x}={k\over k-x}-1$, it follows, on taking derivatives, that

$$\sum_{n=1}^\infty\zeta(n+1)nx^{n-1}=\sum_{k=1}^\infty{1\over(k-x)^2}$$

Consequently

$$\sum_{n=1}^\infty\zeta(n+1){n\over2^n}={1\over2}\sum_{k=1}^\infty{1\over(k-1/2)^2}=2\sum_{k=1}^\infty{1\over(2k-1)^2}=2\left(\sum_{k=1}^\infty{1\over k^2}-\sum_{k=1}^\infty{1\over(2k)^2} \right)={3\over2}\zeta(2)={\pi^2\over4}$$

Answer 2

We have $\sum_{n=1}^{\infty}\zeta(n+1)z^n=-\gamma-\psi(1-z)$ hence $\sum_{n=1}^{\infty}n\zeta(n+1)z^{n-1}=\psi'(1-z)$. Or (the same): $$\sum_{n=1}^{\infty}nz^{n-1}\sum_{k=1}^{\infty}\frac{1}{k^{n+1}}=\sum_{k=1}^{\infty}\frac{1}{k^2}\sum_{n=1}^{\infty}n\left(\frac{z}{k}\right)^{n-1}=\sum_{k=1}^{\infty}\frac{1}{(k-z)^2}.$$ The given sum is then equal to $\psi'(1/2)/2=\pi^2/4$ (which can be obtained using the last sum).

Answer 3

The logarithmic derivative of the gamma function $\psi$ has the power series expansion $$ \psi (1 - z) + \gamma = - \sum\limits_{n = 1}^\infty {\zeta (n + 1)z^n } $$ provided $|z|<1$ (see $(5.7.4)$). Differentiating both sides with respect to $z$ and substituting $z=1/2$, we obtain $$ - \psi '\!\left( {\tfrac{1}{2}} \right) = - 2\sum\limits_{n = 1}^\infty {\zeta (n + 1)\frac{n}{{2^n }}} . $$ Thus, $$ \sum\limits_{n = 1}^\infty {\zeta (n + 1)\frac{n}{{2^n }}} = \frac{1}{2}\psi '\!\left( {\tfrac{1}{2}} \right) = \frac{3}{2}\zeta (2) = \frac{{\pi ^2 }}{4}, $$ where we used $(5.15.3)$.