級数一覧　𝟑

(𝟷)${\displaystyle \sum _{n=1}^{\mathrm{\infty }}\frac{\tanh \pi wn}{wn(n^2-z^2)}+\sum _{n=0}^{\mathrm{\infty }}\frac{w}{(n+\frac{1}{2})({(n+\frac{1}{2})}^2+(wz{)}^2)\tanh \frac{\pi }{w}(n+\frac{1}{2})}=\frac{\pi }{2z^2}(1-\frac{\tanh \pi wz}{w\tan \pi z})}$

(𝟸)${\displaystyle \sum _{n=1}^{\mathrm{\infty }}\frac{1}{(n^2-z^2)\cosh \pi wn}+\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^{n}w}{({(n+\frac{1}{2})}^2+(wz{)}^2)\tanh \frac{\pi }{w}(n+\frac{1}{2})}=\frac{1}{2z^2}(1-\frac{\pi z}{\tan \pi z\cosh \pi wz})}$

(𝟹)${\displaystyle \sum _{n=1}^{\mathrm{\infty }}\frac{1}{n(n^2-z^2)\tanh \pi wn}+\sum _{n=1}^{\mathrm{\infty }}\frac{w^2}{n(n^2+(wz{)}^2)}\frac{1}{\tanh \frac{\pi n}{w}}=\frac{1}{2\pi w{z}^4}-\frac{\pi (1-w^2)}{6w{z}^2}-\frac{\pi }{2z^2\tan \pi z\tanh \pi wz}}$

(𝟺)${\displaystyle \sum _{n=1}^{\mathrm{\infty }}\frac{1}{(n^2-z^2){\cosh }^2\pi n}+\sum _{n=0}^{\mathrm{\infty }}(\frac{1}{({(n+\frac{1}{2})}^2+z^2){\tanh }^2\pi (n+\frac{1}{2})}+\frac{2n+1}{\pi {({(n+\frac{1}{2})}^2+z^2)}^2\tanh \pi (n+\frac{1}{2})})=\frac{\pi \tanh \pi z}{2z}+\frac{1}{2z^2}-\frac{\pi }{2z\tan \pi z{\cosh }^2\pi z}}$

(𝟻)${\displaystyle \sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^{n}w}{({(n+\frac{1}{2})}^2+(wz{)}^2)\sinh \frac{\pi }{w}(n+\frac{1}{2})}-\sum _{n=1}^{\mathrm{\infty }}\frac{(-1{)}^{n-1}}{(n^2-z^2)\cosh \pi wn}=\frac{1}{2z^2}(1-\frac{\pi z}{\sin \pi z\cosh \pi wz})}$

(𝟼)${\displaystyle \sum _{n=1}^{\mathrm{\infty }}\frac{(-1{)}^{n-1}n\tanh \pi wn}{n^2-z^2}+\sum _{n=0}^{\mathrm{\infty }}\frac{n+\frac{1}{2}}{({(n+\frac{1}{2})}^2+(wz{)}^2)\sinh \frac{\pi }{w}(n+\frac{1}{2})}=\frac{\pi }{2}\frac{\tanh \pi wz}{\sin \pi z}}$

(𝟽)${\displaystyle \sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^2-z^2}\frac{\sinh \frac{\pi nq}{w}}{\sinh \frac{\pi n}{w}}+\sum _{n=1}^{\mathrm{\infty }}\frac{(-1{)}^{n-1}w\sin \pi qn}{((wn{)}^2+z^2)\tanh \pi wn}=\frac{q}{2z^2}-\frac{\pi }{2z\tan \pi z}\frac{\sinh \frac{\pi qz}{w}}{\sinh \frac{\pi z}{w}}}$

(𝟾)${\displaystyle \sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^{n-1}}{n^2+z^2}\frac{\sinh \pi qnw}{\sinh \pi nw}-\sum _{n=1}^{\mathrm{\infty }}\frac{w{(-1)}^{n-1}\sin \pi qn}{(n^2-{\left(wz\right)}^2)\sinh \frac{\pi n}{w}}=\frac{q}{2z^2}-\frac{\pi }{2z\sinh \pi z}\frac{\sin \pi qzw}{\sin \pi zw}}$

(𝟿)$\begin{array}{r}{\displaystyle {\mathcal{N}}_{\pm }:=2n+1\pm w}\end{array}$，$\begin{array}{r}{\displaystyle A_r({\mathcal{N}}_{\pm }):=\sum _{k=0}^r\frac{2q^{2k}\zeta (2r-2k)}{\pi {{\mathcal{N}}_{\pm }}^{2k+1}}}\end{array}$と定義すれば

$\begin{array}{r}{\displaystyle \sum _{n=0}^{\mathrm{\infty }}(A_r({\mathcal{N}}_{\mathcal{-}})-A_r({\mathcal{N}}_{\mathcal{+}})-q(\frac{1}{{\mathcal{N}}_{-}^{2r}\tanh \frac{\pi {\mathcal{N}}_{-}}{q}}-\frac{1}{{\mathcal{N}}_{+}^{2r}\tanh \frac{\pi {\mathcal{N}}_{+}}{q}}))=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^{2r}}\frac{\sin \pi w}{\cosh \pi qn+\cos \pi w}}\end{array}$