How do you prove that ${\sum_{n=1}^{k}(\zeta(2*n)/n)-H_k(1)}$ tends to $\ln(2)$ as integer $k$ tends to infinity where $H_k(1) = \sum_{n=1}^{k}{1\over n}$? Is this result well known? Please give a reference if it is. $\zeta(x)$ is the Riemann zeta function evaluated at $x$. I have a proof.
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Riemann zeta function - Infinite series has this series and some others. – Alexander Vlasev Nov 28 '14 at 08:40
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@Aleks Vlasev. That is a great help. Thank you. I found the result by accident by inspecting the infinite product over the integers, n>1 of (1-1/n^2) which is easy to show equals 1/2 [I wanted an example of a monotonic increasing sequence<1 whose product is >0 for other work I was doing] and taking the log of both sides, expanding each term in the resulting infinite sum i.e. log(1-1/n^2) as another infinite series and regrouping the result to give the above formula. – Paul Masham Nov 28 '14 at 18:58