無限和

$$\begin{array}{rlrl}& (1)& & \sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^2}=\zeta (2)={\mathrm{L}\mathrm{i}}_2(1)=\frac{{\pi }^2}{6}\\ & (2)& & \sum _{n=1}^{\mathrm{\infty }}\frac{1}{(2n{)}^2}=\frac{1}{4}\zeta (2)\\ & (3)& & \sum _{n=1}^{\mathrm{\infty }}\frac{1}{(2n-1{)}^2}=\frac{3}{4}\zeta (2)\\ & (4)& & \sum _{n=1}^{\mathrm{\infty }}\frac{(-1{)}^n}{n^2}={\mathrm{L}\mathrm{i}}_2(-1)=-\frac{1}{2}\zeta (2)\\ & (5)& & \sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^3}=\zeta (3)={\mathrm{L}\mathrm{i}}_3(1)\\ & (6)& & \sum _{n=1}^{\mathrm{\infty }}\frac{1}{(2n{)}^3}=\frac{1}{8}\zeta (3)\\ & (7)& & \sum _{n=1}^{\mathrm{\infty }}\frac{1}{(2n-1{)}^3}=\frac{7}{8}\zeta (3)\\ & (8)& & \sum _{n=1}^{\mathrm{\infty }}\frac{(-1{)}^n}{n^3}={\mathrm{L}\mathrm{i}}_3(-1)=-\frac{3}{4}\zeta (3)\\ & (9)& & \sum _{n=1}^{\mathrm{\infty }}\frac{1}{(3n{)}^3}=\frac{1}{27}\zeta (3)\\ & (10)& & \sum _{n=1}^{\mathrm{\infty }}\frac{1}{(3n-2{)}^3}=\frac{13}{27}\zeta (3)+\frac{2\sqrt{3}{\pi }^3}{243}\\ & (11)& & \sum _{n=1}^{\mathrm{\infty }}\frac{1}{(3n-1{)}^3}=\frac{13}{27}\zeta (3)-\frac{2\sqrt{3}{\pi }^3}{243}\\ & (12)& & \sum _{n=1}^{\mathrm{\infty }}\frac{{\omega }^n}{n^3}={\mathrm{L}\mathrm{i}}_3(\omega )=-\frac{4}{9}\zeta (3)+\frac{2{\pi }^3}{81}i\\ & (13)& & \sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^4}=\zeta (4)={\mathrm{L}\mathrm{i}}_4(1)=\frac{{\pi }^4}{90}\\ & (14)& & \sum _{n=1}^{\mathrm{\infty }}\frac{1}{(2n{)}^4}=\frac{1}{16}\zeta (4)\\ & (15)& & \sum _{n=1}^{\mathrm{\infty }}\frac{1}{(2n-1{)}^4}=\frac{15}{16}\zeta (4)\\ & (16)& & \sum _{n=1}^{\mathrm{\infty }}\frac{(-1{)}^n}{n^4}={\mathrm{L}\mathrm{i}}_4(-1)=-\frac{7}{8}\zeta (4)\\ & (17)& & \sum _{n=1}^{\mathrm{\infty }}\frac{1}{(3n{)}^4}=\frac{1}{81}\zeta (4)\\ & (18)& & \sum _{n=1}^{\mathrm{\infty }}\frac{1}{(3n-2{)}^4}=\frac{1}{486}{\psi }^{(3)}\left(\frac{1}{3}\right)\\ & (19)& & \sum _{n=1}^{\mathrm{\infty }}\frac{1}{(3n-1{)}^4}=\frac{1}{486}{\psi }^{(3)}\left(\frac{2}{3}\right)\\ & (20)& & \sum _{n=1}^{\mathrm{\infty }}\frac{{\omega }^n}{n^4}={\mathrm{L}\mathrm{i}}_4(\omega )=-\frac{79}{81}\zeta (4)+\frac{\sqrt{3}}{972}({\psi }^{(3)}\left(\frac{1}{3}\right)-{\psi }^{(3)}\left(\frac{2}{3}\right))i\\ & (21)& & \sum _{n=1}^{\mathrm{\infty }}\frac{1}{(4n{)}^4}=\frac{1}{256}\zeta (4)\\ & (22)& & \sum _{n=1}^{\mathrm{\infty }}\frac{1}{(4n-3{)}^4}=\frac{1}{1536}{\psi }^{(3)}\left(\frac{1}{4}\right)\\ & (23)& & \sum _{n=1}^{\mathrm{\infty }}\frac{1}{(4n-2{)}^4}=\frac{15}{256}\zeta (4)\\ & (24)& & \sum _{n=1}^{\mathrm{\infty }}\frac{1}{(4n-1{)}^4}=\frac{1}{1536}{\psi }^{(3)}\left(\frac{3}{4}\right)\\ & (25)& & \sum _{n=1}^{\mathrm{\infty }}\frac{i^n}{n^4}={\mathrm{L}\mathrm{i}}_4(i)=-\frac{7}{128}\zeta (4)+\frac{1}{1536}({\psi }^{(3)}\left(\frac{1}{4}\right)-{\psi }^{(3)}\left(\frac{3}{4}\right))i\end{array}$$
ただし、$i$を虚数単位、$\pi$を円周率、${\displaystyle \omega :=\frac{-1+\sqrt{3}i}{2}}$、${\displaystyle {\psi }^{(3)}(z)}$をペンタガンマ函数、${\displaystyle {\mathrm{L}\mathrm{i}}_s(z)}$を多重対数函数、${\displaystyle \zeta (s)}$をRiemannのゼータ函数とする。