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Question Number 63665 by mathmax by abdo last updated on 07/Jul/19

find the value of Σ_(n=1) ^∞ (((−1)^n )/(n^2 (n+1)^3 ))

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

Commented by mathmax by abdo last updated on 07/Jul/19

let decompose F(x)=(1/(x^2 (x+1)^3 )) ⇒ F(x)=(a/x) +(b/x^2 ) +(c/(x+1)) +(d/((x+1)^2 )) +(e/((x+1)^3 )) b=lim_(x→0) x^2 F(x)=1 e =lim_(x→−1) (x+1)^3 F(x)=1 ⇒F(x)=(a/x) +(1/x^2 ) +(c/(x+1)) +(d/((x+1)^2 )) +(1/((x+1)^3 )) F(x)−(1/x^2 ) =(1/x^2 ){(1/((x+1)^3 )) −1} =(1/x^2 ){((1−x^3 −3x^2 −3x−1)/((x+1)^3 ))} =−((x^3 +3x^2 +3x)/(x^2 (x+1)^3 )) =−((x^2 +3x +3)/(x(x+1)^3 )) =(a/x) +(c/(x+1)) +(d/((x+1)^2 )) +(1/((x+1)^3 )) ⇒ −((x^2 +3x +3)/((x+1)^3 )) =a +((cx)/(x+1)) +(dx/((x+1)^2 )) +(x/((x+1)^3 )) (x→0) ⇒ a =−3 ⇒F(x) =−(3/x) +(1/x^2 ) +(c/(x+1)) +(d/((x+1)^2 )) +(1/((x+1)^3 )) lim_(x→+∞) xF(x)=0 =−3 +c ⇒c=3 ⇒ F(x)=−(3/x) +(1/x^2 ) +(3/(x+1)) +(d/((x+1)^2 )) +(1/((x+1)^3 )) F(−2) =−(1/4) =(3/2) +(1/4) −3 +d −1 =(7/4)−4 +d ⇒ d =−(1/4)−(7/4) +4 =−2+4 =2 ⇒ F(x)=((−3)/x) +(1/x^2 ) +(3/(x+1)) +(2/((x+1)^2 )) +(1/((x+1)^3 )) S =Σ_(n=1) ^∞ (−1)^n F(n) =−3 Σ_(n=1) ^∞ (((−1)^n )/n) +Σ_(n=1) ^∞ (((−1)^n )/n^2 ) +2Σ_(n=1) ^∞ (((−1)^n )/((n+1)^2 )) +Σ_(n=1) ^∞ (((−1)^n )/((n+1)^3 )) we have Σ_(n=1) ^∞ (x^n /n) =−ln(1−x) if ∣x∣<1 ⇒Σ_(n=1) ^∞ (((−1)^n )/n) =−ln(2) if δ(x) =Σ_(n=1) ^∞ (((−1)^n )/n^x ) we have proved that δ(x)=(2^(1−x) −1)ξ(x) ⇒Σ_(n=1) ^∞ (((−1)^n )/n^2 ) =(2^(−1) −1)ξ(2) =−(1/2)(π^2 /6) =−(π^2 /(12)) Σ_(n=1) ^∞ (((−1)^n )/((n+1)^2 )) =Σ_(n=2) ^∞ (((−1)^(n−1) )/n^2 ) =−Σ_(n=2) ^∞ (((−1)^n )/n^2 ) =−{−(π^2 /(12)) +1}=(π^2 /(12)) −1 also Σ_(n=1) ^∞ (((−1)^n )/((n+1)^3 )) =Σ_(n=2) ^∞ (((−1)^(n−1) )/n^3 ) =−Σ_(n=2) ^∞ (((−1)^n )/n^3 ) =−{ δ(3) +1} =−(2^(−2) −1)ξ(3)−1 =−(−(3/4))ξ(3)−1 =(3/4)ξ(3)−1 ⇒ S = 3ln(2)−(π^2 /(12)) +((2π^2 )/(12)) −2 +(3/4)ξ(3)−1 =3ln(2) +(π^2 /(12)) +(3/4)ξ(3)−3 .

$${let}\:{decompose}\:{F}\left({x}\right)=\frac{\mathrm{1}}{{x}^{\mathrm{2}} \left({x}+\mathrm{1}\right)^{\mathrm{3}} }\:\Rightarrow \\ $$$${F}\left({x}\right)=\frac{{a}}{{x}}\:+\frac{{b}}{{x}^{\mathrm{2}} }\:+\frac{{c}}{{x}+\mathrm{1}}\:+\frac{{d}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }\:+\frac{{e}}{\left({x}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$$${b}={lim}_{{x}\rightarrow\mathrm{0}} {x}^{\mathrm{2}} {F}\left({x}\right)=\mathrm{1} \\ $$$${e}\:={lim}_{{x}\rightarrow−\mathrm{1}} \left({x}+\mathrm{1}\right)^{\mathrm{3}} {F}\left({x}\right)=\mathrm{1}\:\Rightarrow{F}\left({x}\right)=\frac{{a}}{{x}}\:+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:+\frac{{c}}{{x}+\mathrm{1}}\:+\frac{{d}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }\:+\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$$${F}\left({x}\right)−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:=\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\left\{\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)^{\mathrm{3}} }\:−\mathrm{1}\right\}\:=\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\left\{\frac{\mathrm{1}−{x}^{\mathrm{3}} −\mathrm{3}{x}^{\mathrm{2}} −\mathrm{3}{x}−\mathrm{1}}{\left({x}+\mathrm{1}\right)^{\mathrm{3}} }\right\} \\ $$$$=−\frac{{x}^{\mathrm{3}} \:+\mathrm{3}{x}^{\mathrm{2}} +\mathrm{3}{x}}{{x}^{\mathrm{2}} \left({x}+\mathrm{1}\right)^{\mathrm{3}} }\:=−\frac{{x}^{\mathrm{2}} \:+\mathrm{3}{x}\:+\mathrm{3}}{{x}\left({x}+\mathrm{1}\right)^{\mathrm{3}} }\:=\frac{{a}}{{x}}\:+\frac{{c}}{{x}+\mathrm{1}}\:+\frac{{d}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }\:+\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)^{\mathrm{3}} }\:\Rightarrow \\ $$$$−\frac{{x}^{\mathrm{2}} +\mathrm{3}{x}\:+\mathrm{3}}{\left({x}+\mathrm{1}\right)^{\mathrm{3}} }\:={a}\:+\frac{{cx}}{{x}+\mathrm{1}}\:+\frac{{dx}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }\:+\frac{{x}}{\left({x}+\mathrm{1}\right)^{\mathrm{3}} }\:\left({x}\rightarrow\mathrm{0}\right)\:\Rightarrow \\ $$$${a}\:=−\mathrm{3}\:\Rightarrow{F}\left({x}\right)\:=−\frac{\mathrm{3}}{{x}}\:+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:+\frac{{c}}{{x}+\mathrm{1}}\:+\frac{{d}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }\:+\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$$${lim}_{{x}\rightarrow+\infty} {xF}\left({x}\right)=\mathrm{0}\:=−\mathrm{3}\:+{c}\:\Rightarrow{c}=\mathrm{3}\:\Rightarrow \\ $$$${F}\left({x}\right)=−\frac{\mathrm{3}}{{x}}\:+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:+\frac{\mathrm{3}}{{x}+\mathrm{1}}\:+\frac{\mathrm{d}}{\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} }\:+\frac{\mathrm{1}}{\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$$$\mathrm{F}\left(−\mathrm{2}\right)\:=−\frac{\mathrm{1}}{\mathrm{4}}\:=\frac{\mathrm{3}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{4}}\:−\mathrm{3}\:+{d}\:−\mathrm{1}\:=\frac{\mathrm{7}}{\mathrm{4}}−\mathrm{4}\:+{d}\:\Rightarrow \\ $$$${d}\:=−\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{7}}{\mathrm{4}}\:+\mathrm{4}\:=−\mathrm{2}+\mathrm{4}\:=\mathrm{2}\:\Rightarrow \\ $$$${F}\left({x}\right)=\frac{−\mathrm{3}}{{x}}\:+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:+\frac{\mathrm{3}}{{x}+\mathrm{1}}\:+\frac{\mathrm{2}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }\:+\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$$${S}\:=\sum_{{n}=\mathrm{1}} ^{\infty} \left(−\mathrm{1}\right)^{{n}} {F}\left({n}\right)\:=−\mathrm{3}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}}\:+\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} } \\ $$$$+\mathrm{2}\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\:+\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left({n}+\mathrm{1}\right)^{\mathrm{3}} }\:\:{we}\:{have}\: \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{x}^{{n}} }{{n}}\:=−{ln}\left(\mathrm{1}−{x}\right)\:\:{if}\:\mid{x}\mid<\mathrm{1}\:\Rightarrow\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}}\:=−{ln}\left(\mathrm{2}\right) \\ $$$${if}\:\delta\left({x}\right)\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{{x}} }\:\:\:\:{we}\:{have}\:{proved}\:{that}\:\delta\left({x}\right)=\left(\mathrm{2}^{\mathrm{1}−{x}} −\mathrm{1}\right)\xi\left({x}\right) \\ $$$$\Rightarrow\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} }\:=\left(\mathrm{2}^{−\mathrm{1}} −\mathrm{1}\right)\xi\left(\mathrm{2}\right)\:=−\frac{\mathrm{1}}{\mathrm{2}}\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:=−\frac{\pi^{\mathrm{2}} }{\mathrm{12}} \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\:=\sum_{{n}=\mathrm{2}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}^{\mathrm{2}} }\:=−\sum_{{n}=\mathrm{2}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} } \\ $$$$=−\left\{−\frac{\pi^{\mathrm{2}} }{\mathrm{12}}\:+\mathrm{1}\right\}=\frac{\pi^{\mathrm{2}} }{\mathrm{12}}\:−\mathrm{1}\:\:{also} \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left({n}+\mathrm{1}\right)^{\mathrm{3}} }\:=\sum_{{n}=\mathrm{2}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}^{\mathrm{3}} }\:=−\sum_{{n}=\mathrm{2}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{3}} } \\ $$$$=−\left\{\:\delta\left(\mathrm{3}\right)\:+\mathrm{1}\right\}\:=−\left(\mathrm{2}^{−\mathrm{2}} −\mathrm{1}\right)\xi\left(\mathrm{3}\right)−\mathrm{1}\:=−\left(−\frac{\mathrm{3}}{\mathrm{4}}\right)\xi\left(\mathrm{3}\right)−\mathrm{1} \\ $$$$=\frac{\mathrm{3}}{\mathrm{4}}\xi\left(\mathrm{3}\right)−\mathrm{1}\:\Rightarrow \\ $$$${S}\:=\:\mathrm{3}{ln}\left(\mathrm{2}\right)−\frac{\pi^{\mathrm{2}} }{\mathrm{12}}\:+\frac{\mathrm{2}\pi^{\mathrm{2}} }{\mathrm{12}}\:−\mathrm{2}\:+\frac{\mathrm{3}}{\mathrm{4}}\xi\left(\mathrm{3}\right)−\mathrm{1} \\ $$$$=\mathrm{3}{ln}\left(\mathrm{2}\right)\:+\frac{\pi^{\mathrm{2}} }{\mathrm{12}}\:+\frac{\mathrm{3}}{\mathrm{4}}\xi\left(\mathrm{3}\right)−\mathrm{3}\:. \\ $$$$ \\ $$$$ \\ $$

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