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Question Number 74887 by abdomathmax last updated on 03/Dec/19

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

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

Commented by mathmax by abdo last updated on 21/Dec/19

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

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

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