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Question Number 86958 by abdomathmax last updated on 01/Apr/20

calculate Σ_(n=1) ^∞  (((−1)^n )/(n^3 (n+1)^2 ))

$${calculate}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{3}} \left({n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$

Commented by mathmax by abdo last updated on 04/Apr/20

let decompose F(x)=(1/(x^3 (x+1)^2 )) ⇒  F(x)=(a/x)+(b/x^2 ) +(c/x^3 ) +(d/(x+1)) +(e/((x+1)^2 ))  c=1  and e =−1 ⇒ F(x)=(a/x) +(b/x^2 )+(1/x^3 )+(d/(x+1))−(1/((x+1)^2 ))  lim_(x→+∞)  xF(x)=0 =a+d ⇒d=−a ⇒  F(x)=(a/x) +(b/x^2 ) +(1/x^3 )−(a/(x+1))−(1/((x+1)^2 ))  F(1) =(1/4) =a+b+1−(a/2)−(1/4) ⇒1 =4a+4b−2a−1 ⇒  2a+4b =2 ⇒a +2b =1  F(−2) =−(1/8) =−(a/2)+(b/4)−(1/8) +a−1 ⇒  −1 =−4a +2b−1+8a−8 ⇒0 =4a+2b −8 ⇒  2a+b−4 =0 ⇒2(1−2b)+b−4 =0 ⇒2−4b+b−4 =0 ⇒  −2b−2 =0 ⇒b=−1 ⇒a =1−2(−1) =1 ⇒  F(x)=(1/x)−(1/x^2 )+(1/x^3 )−(1/(x+1))−(1/((x+1)^2 )) ⇒  S =Σ_(n=1) ^∞  (−1)^n F(n)   =Σ_(n=1) ^∞  (((−1)^n )/n)−Σ_(n=1) ^∞  (((−1)^n )/n^2 ) +Σ_(n=1) ^∞  (((−1)^n )/n^3 )−Σ_(n=1) ^∞  (((−1)^n )/(n+1))  −Σ_(n=1) ^∞  (((−1)^n )/((n+1)^2 ))  we have  Σ_(n=1) ^∞  (((−1)^n )/n) =−ln(2)  let δ(x)=Σ_(n=1) ^∞  (((−1)^n )/n^x )  we have δ(x) =(2^(1−x) −1)ξ(x)  Σ_(n=1) ^∞  (((−1)^n )/n^2 ) =δ(2)=(2^(−1) −1)ξ(2) =−(1/2)×(π^2 /6) =−(π^2 /(12))  Σ_(n=1) ^∞  (((−1)^n )/n^3 ) =(2^(1−3) −1)ξ(3) =((1/4)−1)ξ(3) =−(3/4)ξ(3)  Σ_(n=1) ^∞  (((−1)^n )/(n+1)) =Σ_(n=2) ^∞  (((−1)^(n−1) )/n) =−Σ_(n=2) ^∞  (((−1)^n )/n)  =−(Σ_(n=1) ^∞  (((−1)^n )/n) +1) =−(−ln(2)+1) =ln(2)−1  Σ_(n=1) ^∞  (((−1)^n )/((n+1)^2 )) =Σ_(n=2) ^∞  (((−1)^(n−1) )/n^2 ) =−Σ_(n=2) ^∞  (((−1)^n )/n^2 )  =−( Σ_(n=1) ^∞  (((−1)^n )/n^2 ) +1) =−(−(π^2 /(12))+1) =(π^2 /(12))−1 ⇒  S =−ln(2) +(π^2 /(12))−(3/4)ξ(3)−ln(2)+1 −(π^2 /(12)) +1 ⇒  S =2−2ln(2)−(3/4)ξ(3)

$${let}\:{decompose}\:{F}\left({x}\right)=\frac{\mathrm{1}}{{x}^{\mathrm{3}} \left({x}+\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$${F}\left({x}\right)=\frac{{a}}{{x}}+\frac{{b}}{{x}^{\mathrm{2}} }\:+\frac{{c}}{{x}^{\mathrm{3}} }\:+\frac{{d}}{{x}+\mathrm{1}}\:+\frac{{e}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${c}=\mathrm{1}\:\:{and}\:{e}\:=−\mathrm{1}\:\Rightarrow\:{F}\left({x}\right)=\frac{{a}}{{x}}\:+\frac{{b}}{{x}^{\mathrm{2}} }+\frac{\mathrm{1}}{{x}^{\mathrm{3}} }+\frac{{d}}{{x}+\mathrm{1}}−\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${lim}_{{x}\rightarrow+\infty} \:{xF}\left({x}\right)=\mathrm{0}\:={a}+{d}\:\Rightarrow{d}=−{a}\:\Rightarrow \\ $$$${F}\left({x}\right)=\frac{{a}}{{x}}\:+\frac{{b}}{{x}^{\mathrm{2}} }\:+\frac{\mathrm{1}}{{x}^{\mathrm{3}} }−\frac{{a}}{{x}+\mathrm{1}}−\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${F}\left(\mathrm{1}\right)\:=\frac{\mathrm{1}}{\mathrm{4}}\:={a}+{b}+\mathrm{1}−\frac{{a}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{4}}\:\Rightarrow\mathrm{1}\:=\mathrm{4}{a}+\mathrm{4}{b}−\mathrm{2}{a}−\mathrm{1}\:\Rightarrow \\ $$$$\mathrm{2}{a}+\mathrm{4}{b}\:=\mathrm{2}\:\Rightarrow{a}\:+\mathrm{2}{b}\:=\mathrm{1} \\ $$$${F}\left(−\mathrm{2}\right)\:=−\frac{\mathrm{1}}{\mathrm{8}}\:=−\frac{{a}}{\mathrm{2}}+\frac{{b}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{8}}\:+{a}−\mathrm{1}\:\Rightarrow \\ $$$$−\mathrm{1}\:=−\mathrm{4}{a}\:+\mathrm{2}{b}−\mathrm{1}+\mathrm{8}{a}−\mathrm{8}\:\Rightarrow\mathrm{0}\:=\mathrm{4}{a}+\mathrm{2}{b}\:−\mathrm{8}\:\Rightarrow \\ $$$$\mathrm{2}{a}+{b}−\mathrm{4}\:=\mathrm{0}\:\Rightarrow\mathrm{2}\left(\mathrm{1}−\mathrm{2}{b}\right)+{b}−\mathrm{4}\:=\mathrm{0}\:\Rightarrow\mathrm{2}−\mathrm{4}{b}+{b}−\mathrm{4}\:=\mathrm{0}\:\Rightarrow \\ $$$$−\mathrm{2}{b}−\mathrm{2}\:=\mathrm{0}\:\Rightarrow{b}=−\mathrm{1}\:\Rightarrow{a}\:=\mathrm{1}−\mathrm{2}\left(−\mathrm{1}\right)\:=\mathrm{1}\:\Rightarrow \\ $$$${F}\left({x}\right)=\frac{\mathrm{1}}{{x}}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }+\frac{\mathrm{1}}{{x}^{\mathrm{3}} }−\frac{\mathrm{1}}{{x}+\mathrm{1}}−\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$${S}\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\left(−\mathrm{1}\right)^{{n}} {F}\left({n}\right)\: \\ $$$$=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}}−\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} }\:+\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{3}} }−\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}+\mathrm{1}} \\ $$$$−\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\:\:{we}\:{have} \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}}\:=−{ln}\left(\mathrm{2}\right) \\ $$$${let}\:\delta\left({x}\right)=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{{x}} }\:\:{we}\:{have}\:\delta\left({x}\right)\:=\left(\mathrm{2}^{\mathrm{1}−{x}} −\mathrm{1}\right)\xi\left({x}\right) \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} }\:=\delta\left(\mathrm{2}\right)=\left(\mathrm{2}^{−\mathrm{1}} −\mathrm{1}\right)\xi\left(\mathrm{2}\right)\:=−\frac{\mathrm{1}}{\mathrm{2}}×\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:=−\frac{\pi^{\mathrm{2}} }{\mathrm{12}} \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{3}} }\:=\left(\mathrm{2}^{\mathrm{1}−\mathrm{3}} −\mathrm{1}\right)\xi\left(\mathrm{3}\right)\:=\left(\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{1}\right)\xi\left(\mathrm{3}\right)\:=−\frac{\mathrm{3}}{\mathrm{4}}\xi\left(\mathrm{3}\right) \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}+\mathrm{1}}\:=\sum_{{n}=\mathrm{2}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}}\:=−\sum_{{n}=\mathrm{2}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}} \\ $$$$=−\left(\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}}\:+\mathrm{1}\right)\:=−\left(−{ln}\left(\mathrm{2}\right)+\mathrm{1}\right)\:={ln}\left(\mathrm{2}\right)−\mathrm{1} \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\:=\sum_{{n}=\mathrm{2}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}^{\mathrm{2}} }\:=−\sum_{{n}=\mathrm{2}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} } \\ $$$$=−\left(\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} }\:+\mathrm{1}\right)\:=−\left(−\frac{\pi^{\mathrm{2}} }{\mathrm{12}}+\mathrm{1}\right)\:=\frac{\pi^{\mathrm{2}} }{\mathrm{12}}−\mathrm{1}\:\Rightarrow \\ $$$${S}\:=−{ln}\left(\mathrm{2}\right)\:+\frac{\pi^{\mathrm{2}} }{\mathrm{12}}−\frac{\mathrm{3}}{\mathrm{4}}\xi\left(\mathrm{3}\right)−{ln}\left(\mathrm{2}\right)+\mathrm{1}\:−\frac{\pi^{\mathrm{2}} }{\mathrm{12}}\:+\mathrm{1}\:\Rightarrow \\ $$$${S}\:=\mathrm{2}−\mathrm{2}{ln}\left(\mathrm{2}\right)−\frac{\mathrm{3}}{\mathrm{4}}\xi\left(\mathrm{3}\right) \\ $$

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