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Can someone explain why

$$\sum_{k=1}^\infty (ζ[2k+1]-1)=\frac{1}{4}?$$

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  • $\begingroup$ Why, that's simple. Expand every zeta, then swap the order of summation. $\endgroup$ – Ivan Neretin Sep 12 '15 at 13:45
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We can rearrange $$ \begin{align}\sum_{k=1}^\infty(\zeta(2k+1)-1)&=\sum_{k=1}^\infty\sum_{n=2}^\infty\frac1{n^{2k+1}}\\ &=\sum_{n=2}^\infty\sum_{k=1}^\infty\frac1{n^{2k+1}}\\ &=\sum_{n=2}^\infty\frac1{n(n^2-1)}\\ &=\sum_{n=2}^\infty\left(\frac1{2n(n-1)}-\frac1{2(n+1)n}\right)\\&=\frac14\end{align}$$

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Hint. You are allowed to switch the order of summation $$ \begin{align} \sum_{k=1}^\infty (ζ(2k+1)-1)&=\sum_{k=1}^\infty \left(\sum_{n=1}^\infty\frac1{n^{2k+1}}-1\right)\\\\ &=\sum_{k=1}^\infty \sum_{n=2}^\infty\frac1{n^{2k+1}}\\\\ &=\sum_{n=2}^\infty \sum_{k=1}^\infty\frac1{n^{2k+1}}\\\\ &=\sum_{n=2}^\infty \frac1{n(n-1)(n+1)} \end{align} $$ Then you may conclude by a partial fraction decomposition and a telescoping sum.

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Another approach. Since: $$\zeta(2k+1) = \int_{0}^{+\infty}\frac{x^{2k}}{(2k)!}\cdot\frac{1}{e^x-1}\,dx,\qquad 1=\int_{0}^{+\infty}\frac{x^{2k}}{(2k)!}\cdot\frac{1}{e^x}\,dx\tag{1}$$ we have: $$ \begin{eqnarray*}\sum_{k\geq 1}\left(\zeta(2k+1)-1\right)&=&\int_{0}^{+\infty}\frac{1}{e^{2x}-e^{x}}\sum_{k\geq 1}\frac{x^{2k}}{(2k)!}\,dx \\&=&\int_{0}^{+\infty}\frac{\cosh(x)-1}{e^{2x}-e^x}\,dx\\&=&\frac{1}{2}\int_{0}^{+\infty}\left(e^{-x}-e^{-2x}\right)\,dx\\&=&\frac{1}{2}\cdot\left(1-\frac{1}{2}\right)=\color{red}{\frac{1}{4}}.\tag{2}\end{eqnarray*}$$

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