5
$\begingroup$

I want to find a closed form for the following infinite sum: $$\sum_{k=2}^{\infty} \frac{(-1)^k\cdot(k-1)}{k\cdot(k+1)}\cdot \zeta(k)$$ Is it possible? My approach was to transform it into a double series and then to rewrite it in terms of logarithms, but i didn't get something useful.

Thanks!

$\endgroup$
3
$\begingroup$

We have

$$ \sum_{k=2}^{\infty} \frac{(-1)^k\cdot(k-1)}{k\cdot(k+1)}\cdot \zeta(k) = 2 - \log (2\pi). \tag1 $$

Proof. Let $0<x<1$. Observe that $$ \sum_{k=2}^{\infty} \frac{(-1)^k}{k} x^k = x- \log(1+x) $$ gives $$ \sum_{k=2}^{\infty} \frac{(-1)^k\cdot(k-1)}{k\cdot(k+1)}\cdot x^k = -2+\left( \frac 2x +1\right) \log(1+x). \tag2 $$ Using $$ \zeta(k)=1+\sum_{p=2}^{\infty} \frac{1}{p^k}, \quad k\geq 2,$$ then interverting the two summations in the initial series leads to

$$ \sum_{k=2}^{\infty} \frac{(-1)^k\cdot(k-1)}{k\cdot(k+1)}\cdot \zeta(k) =\sum_{p=1}^{\infty} \left( -2+\left( 2p +1\right) \log\left(1+\frac1p\right)\right). \tag3 $$

For $N\geq 2$, we may rewrite the finite sum $$ \begin{align} \sum_{p=1}^{N} \left( -2+\left( 2p +1\right) \log\left(1+\frac1p\right)\right)& = \sum_{p=1}^{N} \left(\left(2p+1\right)\log\left(p+1\right)-\left(2p-1\right)\log p-2\log p-2\right) \\\\ & = \left(2N+1\right)\log\left(N+1\right)-2\log\left(N!\right)-2N \\\\ & =2 - \log (2\pi)+\mathcal{O}\left(\frac 1N\right) \end{align} $$ giving $(1)$ as $N$ tends to $+\infty$, where we have used Stirling's formula.

$\endgroup$
  • 1
    $\begingroup$ Thank you, this was what I mend with expressing it in terms of logarithms, but I wasn't able to do so. Very nice! $\endgroup$ – Redundant Aunt Sep 27 '14 at 10:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.