無限和

\begin{align} &(1)&&\sum_{n=1}^\infty \frac{1}{n^2}=\zeta(2)=\Li_2(1)=\frac{\pi^2}{6} \\ &(2)&&\sum_{n=1}^\infty \frac{1}{(2n)^2}=\frac{1}{4}\zeta(2) \\ &(3)&&\sum_{n=1}^\infty \frac{1}{(2n-1)^2}=\frac{3}{4}\zeta(2) \\ &(4)&&\sum_{n=1}^\infty \frac{(-1)^n}{n^2}={\rm{Li}}_2(-1)=-\frac{1}{2}\zeta(2) \\ &(5)&&\sum_{n=1}^\infty \frac{1}{n^3}=\zeta(3)=\Li_3(1) \\ &(6)&&\sum_{n=1}^\infty \frac{1}{(2n)^3}=\frac{1}{8}\zeta(3) \\ &(7)&&\sum_{n=1}^\infty \frac{1}{(2n-1)^3}=\frac{7}{8}\zeta(3) \\ &(8)&&\sum_{n=1}^\infty \frac{(-1)^n}{n^3}=\Li_3(-1)=-\frac{3}{4}\zeta(3) \\ &(9)&&\sum_{n=1}^\infty \frac{1}{(3n)^3}=\frac{1}{27}\zeta(3) \\ &(10)&&\sum_{n=1}^\infty \frac{1}{(3n-2)^3}=\frac{13}{27}\zeta(3) +\frac{2\sqrt3\pi^3}{243} \\ &(11)&&\sum_{n=1}^\infty \frac{1}{(3n-1)^3}=\frac{13}{27}\zeta(3) -\frac{2\sqrt3\pi^3}{243} \\ &(12)&&\sum_{n=1}^\infty \frac{\omega^n}{n^3}={\rm{Li}}_3(\omega)=-\frac{4}{9}\zeta(3)+\frac{2\pi^3}{81}i \\ &(13)&&\sum_{n=1}^\infty \frac{1}{n^4}=\zeta(4)=\Li_4(1)=\frac{\pi^4}{90} \\ &(14)&&\sum_{n=1}^\infty \frac{1}{(2n)^4}=\frac{1}{16}\zeta(4) \\ &(15)&&\sum_{n=1}^\infty \frac{1}{(2n-1)^4}=\frac{15}{16}\zeta(4) \\ &(16)&&\sum_{n=1}^\infty \frac{(-1)^n}{n^4}=\Li_4(-1)=-\frac{7}{8}\zeta(4) \\ &(17)&&\sum_{n=1}^\infty \frac{1}{(3n)^4}=\frac{1}{81}\zeta(4) \\ &(18)&&\sum_{n=1}^\infty \frac{1}{(3n-2)^4}=\frac{1}{486}\psi^{(3)}\left(\frac{1}{3}\right) \\ &(19)&&\sum_{n=1}^\infty \frac{1}{(3n-1)^4}=\frac{1}{486}\psi^{(3)}\left(\frac{2}{3}\right) \\ &(20)&&\sum_{n=1}^\infty \frac{\omega^n}{n^4}={\rm{Li}}_4(\omega)=-\frac{79}{81}\zeta(4)+\frac{\sqrt3}{972}\left(\psi^{(3)}\left(\frac{1}{3}\right)-\psi^{(3)}\left(\frac{2}{3}\right)\right)i \\ &(21)&&\sum_{n=1}^\infty \frac{1}{(4n)^4}=\frac{1}{256}\zeta(4) \\ &(22)&&\sum_{n=1}^\infty \frac{1}{(4n-3)^4}=\frac{1}{1536}\psi^{(3)}\left(\frac{1}{4}\right) \\ &(23)&&\sum_{n=1}^\infty \frac{1}{(4n-2)^4}=\frac{15}{256}\zeta(4) \\ &(24)&&\sum_{n=1}^\infty \frac{1}{(4n-1)^4}=\frac{1}{1536}\psi^{(3)}\left(\frac{3}{4}\right) \\ &(25)&&\sum_{n=1}^\infty \frac{i^n}{n^4}=\Li_4(i)=-\frac{7}{128}\zeta(4)+\frac{1}{1536}\left(\psi^{(3)}\left(\frac{1}{4}\right)-\psi^{(3)}\left(\frac{3}{4}\right)\right)i \\ \end{align}
ただし、$i$を虚数単位、$\pi$を円周率、$\displaystyle \omega:=\frac{-1+\sqrt3i}{2}$、$\displaystyle \psi^{(3)}(z)$をペンタガンマ函数、$\displaystyle \Li_s(z)$を多重対数函数、$\displaystyle\zeta(s)$をRiemannのゼータ函数とする。