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偶数ゼータの通常型母関数
\[ \sum_{k=1}^{\infty}\zeta(2k)x^{2k}=\frac{1}{2}\left(1-\pi x\tan^{-1}\left(\pi x\right)\right) \] を使うと、
\begin{align*} \frac{\zeta\left(2\right)}{\pi^{2}}-\frac{\zeta\left(4\right)}{\pi^{4}}+\frac{\zeta\left(6\right)}{\pi^{6}}-\frac{\zeta\left(8\right)}{\pi^{8}}+\cdots & =\sum_{k=1}^{\infty}\left(-1\right)^{k+1}\frac{\zeta\left(2k\right)}{\pi^{2k}}\\ & =-\sum_{k=1}^{\infty}\zeta\left(2k\right)\left(\frac{i}{\pi}\right)^{2k}\\ & =-\frac{1}{2}\left(1-\pi\left(\frac{i}{\pi}\right)\tan^{-1}\left(\pi\left(\frac{i}{\pi}\right)\right)\right)\\ & =-\frac{1}{2}\left(1-i\tan^{-1}\left(i\right)\right)\\ & =-\frac{1}{2}\left(1-\tanh^{-1}\left(1\right)\right)\\ & =\frac{1}{2}\left(\tanh^{-1}\left(1\right)-1\right)\\ & =\frac{1}{2}\left(\frac{e^{1}+e^{-1}}{e^{1}-e^{-1}}-1\right)\\ & =\frac{1}{2}\cdot\frac{2e^{-1}}{e-e^{-1}}\\ & =\frac{1}{e^{2}-1} \end{align*} となる。

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直接計算してみる。
\begin{align*} \frac{\zeta\left(2\right)}{\pi^{2}}-\frac{\zeta\left(4\right)}{\pi^{4}}+\frac{\zeta\left(6\right)}{\pi^{6}}-\frac{\zeta\left(8\right)}{\pi^{8}}+\cdots & =\sum_{k=1}^{\infty}\left(-1\right)^{k+1}\frac{\zeta\left(2k\right)}{\pi^{2k}}\\ & =-\sum_{k=1}^{\infty}\left(-\frac{1}{\pi^{2}}\right)^{k}\sum_{j=1}^{\infty}\frac{1}{j^{2k}}\\ & =-\sum_{j=1}^{\infty}\sum_{k=1}^{\infty}\left(-\frac{1}{\left(\pi j\right)^{2}}\right)^{k}\\ & =-\sum_{j=1}^{\infty}\frac{-\frac{1}{\left(\pi j\right)^{2}}}{1+\frac{1}{\left(\pi j\right)^{2}}}\\ & =\sum_{j=1}^{\infty}\frac{1}{\left(\pi j\right)^{2}+1}\\ & =\frac{1}{\pi^{2}}\sum_{j=1}^{\infty}\frac{1}{j^{2}+\frac{1}{\pi^{2}}}\\ & =\frac{1}{\pi^{2}}\cdot\frac{\pi}{2i}\sum_{j=1}^{\infty}\left(\frac{1}{j-\frac{i}{\pi}}-\frac{1}{j+\frac{i}{\pi}}\right)\\ & =\frac{i}{2\pi}\left(\sum_{j=1}^{\infty}\left(\frac{1}{\frac{i}{\pi}-j}+\frac{1}{\frac{i}{\pi}+j}\right)+\frac{1}{\frac{i}{\pi}+0}-\frac{1}{\frac{i}{\pi}+0}\right)\\ & =\frac{i}{2\pi}\left(\sum_{j=\infty}^{\infty}\frac{1}{\frac{i}{\pi}+j}-\frac{1}{\frac{i}{\pi}+0}\right)\\ & =\frac{i}{2\pi}\left(\pi\tan^{-1}i+i\pi\right)\cmt{\because\pi\tan^{-1}\pi x=\sum_{k=-\infty}^{\infty}\frac{1}{x+k}}\\ & =\frac{i}{2\pi}\left(\frac{\pi}{i}\tanh^{-1}1+i\pi\right)\\ & =\frac{1}{2}\tanh^{-1}1-\frac{1}{2}\\ & =\frac{1}{2}\left(\tanh^{-1}1-1\right)\\ & =\frac{1}{2}\left(\frac{e^{1}+e^{-1}}{e^{1}-e^{-1}}-1\right)\\ & =\frac{1}{2}\cdot\frac{2e^{-1}}{e-e^{-1}}\\ & =\frac{1}{e^{2}-1} \end{align*}