Question Number 159045 by amin96 last updated on 12/Nov/21
$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{3}{n}+\mathrm{1}\right)^{\mathrm{3}} }=? \\ $$
Answered by mindispower last updated on 16/Nov/21
$$=\underset{{n}\geqslant\mathrm{0}} {\sum}.\frac{\mathrm{1}}{\mathrm{27}}\frac{\mathrm{1}}{\left({n}+\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{3}} } \\ $$$$\Psi^{\left(\mathrm{2}\right)} \left({z}\right)=\mathrm{2}\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{\left({n}+{z}\right)^{\mathrm{3}} } \\ $$$${we}\:{get}\:\frac{\mathrm{1}}{\mathrm{54}}\Psi^{\left(\mathrm{2}\right)} \left(\frac{\mathrm{1}}{\mathrm{3}}\right) \\ $$