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Question Number 68593 by Abdo msup. last updated on 14/Sep/19

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

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

Commented by Abdo msup. last updated on 06/Oct/19

let S=Σ_(n=1) ^∞  (((−1)^n )/(n(2n+1)^2 ))  first let decompose  F(x)=(1/(x(2x+1)^2 )) ⇒F(x)=(a/x) +(b/(2x+1)) +(c/((2x+1)^2 ))  a=lim_(x→0) xF(x)=1  c=lim_(x→−(1/2))    (2x+1)^2 F−x)=−2 ⇒  F(x)=(1/x) +(b/(2x+1))−(2/((2x+1)^2 ))  lim_(x→+∞)  xF(x)=0 =1+(b/2) ⇒b+2=0 ⇒b=−2 ⇒  F(x)=(1/x) −(2/(2x+1)) −(2/((2x+1)^2 )) ⇒  S=Σ_(n=1) ^∞  (((−1)^n )/n) −2Σ_(n=1) ^∞  (((−1)^n )/(2n+1)) −2Σ_(n=1) ^∞  (((−1)^n )/((2n+1)^2 ))  we have Σ_(n=1) ^∞  (((−1)^n )/n) =−ln(2)  Σ_(n=1) ^∞  (((−1)^n )/((2n+1))) =Σ_(n=0) ^∞  (((−1)^n )/(2n+1)) −1=(π/4)−1  rest to ca<culate α_0 =Σ_(n=1) ^∞  (((−1)^n )/((2n+1)^2 )) ⇒  S=−ln(2)−(π/2) +2 −2α_0

$${let}\:{S}=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }\:\:{first}\:{let}\:{decompose} \\ $$$${F}\left({x}\right)=\frac{\mathrm{1}}{{x}\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow{F}\left({x}\right)=\frac{{a}}{{x}}\:+\frac{{b}}{\mathrm{2}{x}+\mathrm{1}}\:+\frac{{c}}{\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${a}={lim}_{{x}\rightarrow\mathrm{0}} {xF}\left({x}\right)=\mathrm{1} \\ $$$$\left.{c}={lim}_{{x}\rightarrow−\frac{\mathrm{1}}{\mathrm{2}}} \:\:\:\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{2}} {F}−{x}\right)=−\mathrm{2}\:\Rightarrow \\ $$$${F}\left({x}\right)=\frac{\mathrm{1}}{{x}}\:+\frac{{b}}{\mathrm{2}{x}+\mathrm{1}}−\frac{\mathrm{2}}{\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${lim}_{{x}\rightarrow+\infty} \:{xF}\left({x}\right)=\mathrm{0}\:=\mathrm{1}+\frac{{b}}{\mathrm{2}}\:\Rightarrow{b}+\mathrm{2}=\mathrm{0}\:\Rightarrow{b}=−\mathrm{2}\:\Rightarrow \\ $$$${F}\left({x}\right)=\frac{\mathrm{1}}{{x}}\:−\frac{\mathrm{2}}{\mathrm{2}{x}+\mathrm{1}}\:−\frac{\mathrm{2}}{\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$${S}=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}}\:−\mathrm{2}\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}{n}+\mathrm{1}}\:−\mathrm{2}\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${we}\:{have}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}}\:=−{ln}\left(\mathrm{2}\right) \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\boldsymbol{{n}}} }{\mathrm{2}\boldsymbol{{n}}+\mathrm{1}}\:−\mathrm{1}=\frac{\pi}{\mathrm{4}}−\mathrm{1} \\ $$$${rest}\:{to}\:{ca}<{culate}\:\alpha_{\mathrm{0}} =\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$${S}=−{ln}\left(\mathrm{2}\right)−\frac{\pi}{\mathrm{2}}\:+\mathrm{2}\:−\mathrm{2}\alpha_{\mathrm{0}} \\ $$

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