級数一覧　𝟑

(𝟷)$\qquad\D \sum_{n=1}^\infty\frac{\tanh\pi wn}{wn(n^2-z^2)}+\sum_{n=0}^\infty\frac{w}{\L(n+\frac{1}{2}\R)\L(\L(n+\frac{1}{2}\R)^2+(wz)^2\R)\tanh\frac{\pi}{w}\L(n+\frac{1}{2}\R)}=\frac{\pi}{2z^2}\L(1-\frac{\tanh\pi wz}{w\tan\pi z}\R) $

(𝟸)$\qquad\D \sum_{n=1}^\infty\frac{1}{(n^2-z^2)\cosh\pi wn}+\sum_{n=0}^\infty \frac{(-1)^nw}{\L(\L(n+\frac{1}{2}\R)^2+(wz)^2\R)\tanh\frac{\pi}{w}\L(n+\frac{1}{2}\R)}=\frac{1}{2z^2}\L(1-\frac{\pi z}{\tan\pi z\cosh\pi wz}\R) $

(𝟹)$\qquad\D \sum_{n=1}^\infty\frac{1}{n(n^2-z^2)\tanh\pi wn}+\sum_{n=1}^\infty\frac{w^2}{n(n^2+(wz)^2)}\frac{1}{\tanh\frac{\pi n}{w}} =\frac{1}{2\pi wz^4}-\frac{\pi(1-w^2)}{6wz^2}-\frac{\pi}{2z^2\tan\pi z\tanh\pi wz} $

(𝟺)$\qquad\D \sum_{n=1}^\infty\frac{1}{(n^2-z^2)\cosh^2\pi n}+\sum_{n=0}^\infty\L(\frac{1}{\L(\L(n+\frac{1}{2}\R)^2+z^2\R)\tanh^2\pi\L(n+\frac{1}{2}\R)}+\frac{2n+1}{\pi\L(\L(n+\frac{1}{2}\R)^2+z^2\R)^2\tanh\pi\L(n+\frac{1}{2}\R)}\R) =\frac{\pi\tanh\pi z}{2z}+\frac{1}{2z^2}-\frac{\pi}{2z\tan\pi z\cosh^2\pi z} $

(𝟻)$\qquad\D \sum_{n=0}^\infty \frac{(-1)^nw}{\left(\left(n+\frac{1}{2}\right)^2+(wz)^2\right)\sinh\frac{\pi}{w}\left(n+\frac{1}{2}\right)} -\sum_{n=1}^\infty\frac{(-1)^{n-1}}{\left(n^2-z^2\right)\cosh\pi wn} =\frac{1}{2z^2}\left(1-\frac{\pi z}{\sin\pi z\cosh\pi wz}\right) $

(𝟼)$\qquad\D \sum_{n=1}^\infty \frac{(-1)^{n-1}n\tanh\pi wn}{n^2-z^2}+\sum_{n=0}^\infty \frac{n+\frac{1}{2}}{\left(\left(n+\frac{1}{2}\right)^2+(wz)^2\right)\sinh\frac{\pi}{w}\left(n+\frac{1}{2}\right)} =\frac{\pi}{2}\frac{\tanh\pi wz}{\sin\pi z} $

(𝟽)$\qquad\D \sum_{n=1}^\infty \frac{1}{n^2-z^2}\frac{\sinh\frac{\pi nq}{w}}{\sinh\frac{\pi n}{w}} +\sum_{n=1}^\infty\frac{(-1)^{n-1}w\sin\pi qn}{\L((wn)^2+z^2\R)\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)$\qquad\D \sum_{n=1}^{\infty}\frac{\left(-1\right)^{n-1}}{n^{2}+z^{2}}\frac{\sinh\pi qnw}{\sinh\pi nw}-\sum_{n=1}^{\infty}\frac{w\left(-1\right)^{n-1}\sin\pi qn}{\left(n^{2}-\left(wz\right)^{2}\right)\sinh\frac{\pi n}{w}} =\frac{q}{2z^{2}}-\frac{\pi}{2z\sinh\pi z}\frac{\sin\pi qzw}{\sin\pi zw} $

(𝟿)$\BA\D \mathcal{N_{\pm}}:=2n+1\pm w \EA$，$\BA\D A_{r}({\mathcal{N_{\pm}}}):=\sum_{k=0}^r\frac{2q^{2k}\zeta(2r-2k)}{\pi {\mathcal{N_{\pm}}}^{2k+1}} \EA$と定義すれば

$\qquad\qquad\BA\D \sum_{n=0}^\infty \L(A_{r}({\mathcal{N_{-}}})-A_{r}({\mathcal{N_{+}}})-q\L(\frac{1}{{\mathcal N}_{-}^{2r}\tanh\frac{\pi{\mathcal N}_{-}}{q}}-\frac{1}{{\mathcal N}_{+}^{2r}\tanh\frac{\pi{\mathcal N}_{+}}{q}} \R)\R) =\sum_{n=1}^\infty \frac{1}{n^{2r}}\frac{\sin\pi w}{\cosh\pi qn+\cos\pi w} \EA$