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I've found the following series on the Wikipediapage of the Euler-Mascheroni constant and I want to prove it.

$$\sum_{n=1}^\infty \frac{\zeta(2n+1) - 1}{(2n+1) 2^{2n}} = 1 + \ln(2) - \ln(3) - \gamma$$

I know that $$\sum_{k=1}^\infty \frac{\zeta(2k) - 1}{k} = \ln(2)$$

Any help would be appreciated. Thanks in advance.

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  • $\begingroup$ Please let me know how I can improve my answer. I really want to give you the best answer I can. And feel free to up vote and accept an answer as you see fit. $\endgroup$ – Mark Viola Apr 3 at 3:32
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First, we use the familiar series representation of the Riemann Zeta function to write

$$\zeta(2n+1)-1=\sum_{k=2}^\infty \frac{1}{k^{2n+1}}\tag1$$

Next, define the function $f(x)$ as

$$f(x)=\sum_{n=1}^\infty \frac{\zeta(2n+1)-1}{2n+1}\,x^{2n+1}\tag2$$

Using $(1)$ in $(2)$ reveals

$$\begin{align} f(x)&=\int_0^x \sum_{n=1}^\infty \sum_{k=2}^\infty \frac{t^{2n}}{k^{2n+1}}\,dt\\\\ &=\int_0^x \sum_{k=2}^\infty \frac1k \sum_{n=1}^\infty \left(\frac{t^2}{k^2}\right)^n \,dt \\\\ &=\int_0^x \sum_{k=2}^\infty \frac1k \frac{t^2}{k^2-t^2}\,dt\\\\ &=\int_0^x \left(\frac{t^2}{t^2-1}+\sum_{k=1}^\infty \frac1k\frac{t^2}{k^2-t^2}\right)\,dt\\\\ &=\int_0^x \frac{t^2}{t^2-1}\,dt-\frac12\int_0^x \sum_{k=1}^\infty\left(\frac1k-\frac1{k+t}\right)\,dt-\frac12\int_0^x \sum_{k=1}^\infty\left(\frac1k-\frac1{k-t}\right)\,dt\\\\ &=\int_0^x \frac{t^2}{t^2-1}\,dt-\frac12 \int_0^x \left(2\gamma +\psi(t+1)+\psi(1-t)\right)\,dt\\\\ &=x+\log\left(\frac{1-x}{1+x}\right)-\gamma x -\frac12 \log\left(\frac{\Gamma(1+x)}{\Gamma(1-x)}\right) \end{align}$$

Setting $t=1/2$ and multiplying by $2$ reveals

$$\sum_{n=1}^\infty \frac{\zeta(2n+1)-1}{(2n+1)2^{2n}}=1-\log(3)-\gamma+\log(2)$$

as was to be shown!

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  • $\begingroup$ (+1) Beautiful! $\endgroup$ – mrtaurho Mar 22 at 17:21
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    $\begingroup$ @mrtaurho Thank you! Very much appreciative. $\endgroup$ – Mark Viola Mar 22 at 17:41
  • $\begingroup$ @kevin2806 Please let me know how I can improve my answer. I really want to give you the best answer I can. $\endgroup$ – Mark Viola Mar 25 at 17:46
  • $\begingroup$ And feel free to up vote and accept an answer as you see fit. $\endgroup$ – Mark Viola Mar 25 at 17:46

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