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Question Number 42401 by abdo.msup.com last updated on 24/Aug/18

calculate Σ_(n=0) ^∞ (((−1)^n )/((n+1)^2 ))

$${calculate}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\: \\ $$

Commented by maxmathsup by imad last updated on 25/Aug/18

let S =Σ_(n=0) ^∞ (((−1)^n )/((n+1)^2 )) ⇒S = Σ_(n=1) ^∞ (((−1)^(n−1) )/n^2 ) =Σ_(p=1) ^∞ ((−1)/((2p)^2 )) +Σ_(p=0) ^∞ (1/((2p+1)^2 )) =−(1/4) Σ_(p=1) ^∞ (1/p^2 ) +Σ_(p=0) ^∞ (1/((2p+1)^2 )) but Σ_(p=1) ^∞ (1/p^2 ) =(π^2 /6) Σ_(p=1) ^∞ (1/p^2 ) =Σ_(p=1) ^∞ (1/((2p)^2 )) +Σ_(p=0) ^∞ (1/((2p+1)^2 )) ⇒Σ_(p=0) ^∞ (1/((2p+1)^2 )) =(3/4) (π^2 /6) =(π^2 /8) ⇒ S =−(π^2 /(24)) +(π^2 /8) =((3π^2 −π^2 )/(24)) = (π^2 /(12)) S =(π^2 /(12)) .

$${let}\:{S}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow{S}\:=\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}^{\mathrm{2}} }\:=\sum_{{p}=\mathrm{1}} ^{\infty} \:\:\frac{−\mathrm{1}}{\left(\mathrm{2}{p}\right)^{\mathrm{2}} }\:+\sum_{{p}=\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{1}}{\left(\mathrm{2}{p}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=−\frac{\mathrm{1}}{\mathrm{4}}\:\sum_{{p}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{p}^{\mathrm{2}} }\:+\sum_{{p}=\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{1}}{\left(\mathrm{2}{p}+\mathrm{1}\right)^{\mathrm{2}} }\:{but}\:\:\sum_{{p}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{p}^{\mathrm{2}} }\:=\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$$$\sum_{{p}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{p}^{\mathrm{2}} }\:=\sum_{{p}=\mathrm{1}} ^{\infty} \:\:\frac{\mathrm{1}}{\left(\mathrm{2}{p}\right)^{\mathrm{2}} }\:+\sum_{{p}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{2}{p}+\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow\sum_{{p}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{2}{p}+\mathrm{1}\right)^{\mathrm{2}} }\:=\frac{\mathrm{3}}{\mathrm{4}}\:\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:=\frac{\pi^{\mathrm{2}} }{\mathrm{8}}\:\Rightarrow \\ $$$${S}\:=−\frac{\pi^{\mathrm{2}} }{\mathrm{24}}\:+\frac{\pi^{\mathrm{2}} }{\mathrm{8}}\:=\frac{\mathrm{3}\pi^{\mathrm{2}} −\pi^{\mathrm{2}} }{\mathrm{24}}\:=\:\frac{\pi^{\mathrm{2}} }{\mathrm{12}} \\ $$$${S}\:\:=\frac{\pi^{\mathrm{2}} }{\mathrm{12}}\:. \\ $$

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