Question Number 155133 by amin96 last updated on 25/Sep/21
$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{3}} }=? \\ $$
Answered by mnjuly1970 last updated on 26/Sep/21
$$\:\:\:\frac{\pi^{\:\mathrm{3}} }{\mathrm{32}} \\ $$
Answered by qaz last updated on 26/Sep/21
$$\mathrm{S}=\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\left(\mathrm{2n}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{x}^{\mathrm{2}} \mathrm{e}^{−\mathrm{x}} }{\mathrm{1}+\mathrm{e}^{−\mathrm{2x}} }\mathrm{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\int_{−\infty} ^{\infty} \frac{\mathrm{x}^{\mathrm{2}} \mathrm{e}^{−\mathrm{x}} }{\mathrm{1}+\mathrm{e}^{−\mathrm{2x}} }\mathrm{dx}…….\mathrm{e}^{\mathrm{x}} \rightarrow\mathrm{tan}\:\theta \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \mathrm{ln}^{\mathrm{2}} \mathrm{tan}\:\theta\mathrm{d}\theta \\ $$$$\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \mathrm{tan}^{\mathrm{2k}−\mathrm{1}} \theta\mathrm{d}\theta=\frac{\pi}{\mathrm{2}}\mathrm{csc}\left(\mathrm{k}\pi\right) \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \mathrm{tan}^{\mathrm{2k}−\mathrm{1}} \theta\centerdot\mathrm{ln}^{\mathrm{2}} \mathrm{tan}\:\theta\mathrm{d}\theta=\frac{\pi^{\mathrm{3}} }{\mathrm{8}}\left(\mathrm{csc}\left(\mathrm{k}\pi\right)\mathrm{cot}\:\left(\mathrm{k}\pi\right)+\mathrm{csc}^{\mathrm{3}} \left(\mathrm{k}\pi\right)\right) \\ $$$$\mathrm{let}\:\:\:\mathrm{k}=\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\:\:\:\:\mathrm{we}\:\mathrm{get}\:\:\:\mathrm{S}=\frac{\pi^{\mathrm{3}} }{\mathrm{32}} \\ $$