ζ(2)のような総和
ζ(2)のような総和
次の総和を求めよ。
\[ \sum_{k=0}^{\infty}\frac{1}{k^{2}+1}=? \]
次の総和を求めよ。
\[ \sum_{k=0}^{\infty}\frac{1}{k^{2}+1}=? \]
\begin{align*}
\sum_{k=0}^{\infty}\frac{1}{k^{2}+1} & =\sum_{k=0}^{\infty}\frac{1}{\left(k-i\right)\left(k+i\right)}\\
& =\frac{1}{2i}\sum_{k=0}^{\infty}\left(\frac{1}{k-i}-\frac{1}{k+i}\right)\\
\\
& =\frac{1}{2i}\sum_{k=0}^{\infty}\left\{ \left(\frac{1}{k-i}-\frac{1}{k+1}\right)-\left(\frac{1}{k+i}-\frac{1}{k+1}\right)\right\} \\
& =\frac{1}{2i}\left\{ \left(-\gamma-\psi\left(-i\right)\right)-\left(-\gamma-\psi\left(i\right)\right)\right\} \cmt{\because\psi\left(z\right)=-\gamma-\sum_{k=0}^{\infty}\left(\frac{1}{z+k}-\frac{1}{k+1}\right)}\\
& =\frac{1}{2i}\left(\psi\left(i\right)-\psi\left(-i\right)\right)\\
& =\frac{1}{2i}\left(\psi\left(i\right)-\psi\left(1-i\right)-\frac{1}{i}\right)\cmt{\because\psi\left(z+1\right)=\psi\left(z\right)+\frac{1}{z}}\\
& =-\frac{1}{2i}\left(\psi\left(1-i\right)-\psi\left(i\right)-i\right)\\
& =-\frac{1}{2i}\left(\pi\tan^{-1}\left(\pi i\right)-i\right)\cmt{\because\psi\left(1-z\right)-\psi\left(z\right)=\pi\tan^{-1}\left(\pi z\right)}\\
& =-\frac{1}{2i}\left(\pi i^{-1}\tanh^{-1}\left(\pi\right)-i\right)\\
& =\frac{\pi}{2}\tanh^{-1}\left(\pi\right)+\frac{1}{2}
\end{align*}
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ベータ関数の逆数を含む総和
\[
\lim_{n\rightarrow\infty}\frac{1}{\sqrt{n}}\sum_{k=0}^{n}\frac{\left(-1\right)^{k}}{B\left(n-k+1,k+1\right)\left(2k+1\right)}=?
\]
分母に総和がある数の総和
\[
\frac{1}{1}+\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+\cdots=?
\]
偶数ゼータ関数と円周率を含む交代級数
\[
\frac{\zeta\left(2\right)}{\pi^{2}}-\frac{\zeta\left(4\right)}{\pi^{4}}+\frac{\zeta\left(6\right)}{\pi^{6}}-\frac{\zeta\left(8\right)}{\pi^{8}}+\cdots=?
\]
2項係数の対称性を使います
\[
\sum_{k=0}^{n}kC^{2}\left(n,k\right)=?
\]

