Question Number 42401 by abdo.msup.com last updated on 24/Aug/18
$${calculate}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\: \\ $$
Commented by maxmathsup by imad last updated on 25/Aug/18
$${let}\:{S}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow{S}\:=\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}^{\mathrm{2}} }\:=\sum_{{p}=\mathrm{1}} ^{\infty} \:\:\frac{−\mathrm{1}}{\left(\mathrm{2}{p}\right)^{\mathrm{2}} }\:+\sum_{{p}=\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{1}}{\left(\mathrm{2}{p}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=−\frac{\mathrm{1}}{\mathrm{4}}\:\sum_{{p}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{p}^{\mathrm{2}} }\:+\sum_{{p}=\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{1}}{\left(\mathrm{2}{p}+\mathrm{1}\right)^{\mathrm{2}} }\:{but}\:\:\sum_{{p}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{p}^{\mathrm{2}} }\:=\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$$$\sum_{{p}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{p}^{\mathrm{2}} }\:=\sum_{{p}=\mathrm{1}} ^{\infty} \:\:\frac{\mathrm{1}}{\left(\mathrm{2}{p}\right)^{\mathrm{2}} }\:+\sum_{{p}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{2}{p}+\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow\sum_{{p}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{2}{p}+\mathrm{1}\right)^{\mathrm{2}} }\:=\frac{\mathrm{3}}{\mathrm{4}}\:\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:=\frac{\pi^{\mathrm{2}} }{\mathrm{8}}\:\Rightarrow \\ $$$${S}\:=−\frac{\pi^{\mathrm{2}} }{\mathrm{24}}\:+\frac{\pi^{\mathrm{2}} }{\mathrm{8}}\:=\frac{\mathrm{3}\pi^{\mathrm{2}} −\pi^{\mathrm{2}} }{\mathrm{24}}\:=\:\frac{\pi^{\mathrm{2}} }{\mathrm{12}} \\ $$$${S}\:\:=\frac{\pi^{\mathrm{2}} }{\mathrm{12}}\:. \\ $$