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Question Number 64820 by mathmax by abdo last updated on 22/Jul/19
calculate Σ_(n=1) ^∞ (((−1)^n )/(n^2 (n+1)(n+2)))
$${calculate}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)} \\ $$
Commented by mathmax by abdo last updated on 22/Jul/19
let decompose F(x) =(1/(x^2 (x+1)(x+2))) ⇒ F(x) =(a/x) +(b/x^2 ) +(c/(x+1)) +(d/(x+2)) b=lim_(x→0) x^2 F(x) =(1/2) c =lim_(x→−1) (x+1)F(x) =1 d =lim_(x→−2) (x+2)F(x) =−(1/4) ⇒F(x)=(a/x) +(1/(2x^2 )) +(1/(x+1)) −(1/(4(x+2))) lim_(x→+∞) xF(x) =0 =a+1 ⇒a =−1 ⇒ F(x) =−(1/x) +(1/(2x^2 )) +(1/(x+1)) −(1/(4(x+2))) ⇒ S =−Σ_(n=1) ^∞ (((−1)^n )/n) +(1/2) Σ_(n=1) ^∞ (((−1)^n )/n^2 ) +Σ_(n=1) ^∞ (((−1)^n )/(n+1)) −(1/4)Σ_(n=1) ^∞ (((−1)^n )/(n+2)) we know Σ_(n=1) ^∞ (x^n /n) =−ln(1−x) if ∣x∣<1 ⇒Σ_(n=1) ^∞ (((−1)^n )/n) =−ln2 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) =−(−ln2 +1) =ln2−1 Σ_(n=1) ^∞ (((−1)^n )/(n+2)) =Σ_(n=3) ^(+∞) (((−1)^(n−2) )/n) =Σ_(n=3) ^∞ (((−1)^n )/n) =Σ_(n=1) ^∞ (((−1)^n )/n) −(−1+(1/2)) =−ln2 +(1/2) let δ(x) =Σ_(n=1) ^∞ (((−1)^n )/n^x ) with x>1 we have proved that 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)) ⇒ S =ln(2)−(π^2 /(24)) +ln2−1 −(1/4)(−ln2 +(1/2)) =2ln2 −(π^2 /(24)) −1 +(1/4)ln2 −(1/8) =(9/4)ln2 −(9/8) −(π^2 /(24))
$${let}\:{decompose}\:{F}\left({x}\right)\:=\frac{\mathrm{1}}{{x}^{\mathrm{2}} \left({x}+\mathrm{1}\right)\left({x}+\mathrm{2}\right)}\:\Rightarrow \\ $$$${F}\left({x}\right)\:=\frac{{a}}{{x}}\:+\frac{{b}}{{x}^{\mathrm{2}} }\:+\frac{{c}}{{x}+\mathrm{1}}\:+\frac{{d}}{{x}+\mathrm{2}} \\ $$$${b}={lim}_{{x}\rightarrow\mathrm{0}} {x}^{\mathrm{2}} \:{F}\left({x}\right)\:=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${c}\:={lim}_{{x}\rightarrow−\mathrm{1}} \left({x}+\mathrm{1}\right){F}\left({x}\right)\:=\mathrm{1} \\ $$$${d}\:={lim}_{{x}\rightarrow−\mathrm{2}} \left({x}+\mathrm{2}\right){F}\left({x}\right)\:=−\frac{\mathrm{1}}{\mathrm{4}}\:\Rightarrow{F}\left({x}\right)=\frac{{a}}{{x}}\:+\frac{\mathrm{1}}{\mathrm{2}{x}^{\mathrm{2}} }\:+\frac{\mathrm{1}}{{x}+\mathrm{1}}\:−\frac{\mathrm{1}}{\mathrm{4}\left({x}+\mathrm{2}\right)} \\ $$$${lim}_{{x}\rightarrow+\infty} {xF}\left({x}\right)\:=\mathrm{0}\:={a}+\mathrm{1}\:\Rightarrow{a}\:=−\mathrm{1}\:\Rightarrow \\ $$$${F}\left({x}\right)\:=−\frac{\mathrm{1}}{{x}}\:+\frac{\mathrm{1}}{\mathrm{2}{x}^{\mathrm{2}} }\:+\frac{\mathrm{1}}{{x}+\mathrm{1}}\:−\frac{\mathrm{1}}{\mathrm{4}\left({x}+\mathrm{2}\right)}\:\Rightarrow \\ $$$${S}\:=−\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}}\:+\frac{\mathrm{1}}{\mathrm{2}}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} }\:+\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}+\mathrm{1}}\:−\frac{\mathrm{1}}{\mathrm{4}}\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}+\mathrm{2}} \\ $$$${we}\:{know}\:\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{x}^{{n}} }{{n}}\:=−{ln}\left(\mathrm{1}−{x}\right)\:\:\:{if}\:\mid{x}\mid<\mathrm{1}\:\Rightarrow\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}}\:=−{ln}\mathrm{2} \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}+\mathrm{1}}\:=\sum_{{n}=\mathrm{2}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}}\:=−\sum_{{n}=\mathrm{2}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}} \\ $$$$=−\left(\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}}\:+\mathrm{1}\right)\:=−\left(−{ln}\mathrm{2}\:+\mathrm{1}\right)\:={ln}\mathrm{2}−\mathrm{1} \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}+\mathrm{2}}\:=\sum_{{n}=\mathrm{3}} ^{+\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{2}} }{{n}}\:=\sum_{{n}=\mathrm{3}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}} \\ $$$$=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}}\:−\left(−\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\right)\:=−{ln}\mathrm{2}\:+\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${let}\:\delta\left({x}\right)\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{{x}} }\:\:{with}\:{x}>\mathrm{1}\:\:{we}\:{have}\:{proved}\:{that} \\ $$$$\delta\left({x}\right)\:=\left(\mathrm{2}^{\mathrm{1}−{x}} −\mathrm{1}\right)\xi\left({x}\right)\:\Rightarrow\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} }\:=\delta\left(\mathrm{2}\right)\:=\left(\mathrm{2}^{−\mathrm{1}} −\mathrm{1}\right)\xi\left(\mathrm{2}\right) \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:=−\frac{\pi^{\mathrm{2}} }{\mathrm{12}}\:\:\Rightarrow \\ $$$${S}\:={ln}\left(\mathrm{2}\right)−\frac{\pi^{\mathrm{2}} }{\mathrm{24}}\:+{ln}\mathrm{2}−\mathrm{1}\:−\frac{\mathrm{1}}{\mathrm{4}}\left(−{ln}\mathrm{2}\:+\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$=\mathrm{2}{ln}\mathrm{2}\:−\frac{\pi^{\mathrm{2}} }{\mathrm{24}}\:−\mathrm{1}\:+\frac{\mathrm{1}}{\mathrm{4}}{ln}\mathrm{2}\:−\frac{\mathrm{1}}{\mathrm{8}}\:=\frac{\mathrm{9}}{\mathrm{4}}{ln}\mathrm{2}\:−\frac{\mathrm{9}}{\mathrm{8}}\:−\frac{\pi^{\mathrm{2}} }{\mathrm{24}} \\ $$

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