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Question Number 69376 by mathmax by abdo last updated on 22/Sep/19
calculate Σ_(n=1) ^∞ (((−1)^n )/(n^2 (n+1)^3 ))
$${calculate}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$
Commented by mathmax by abdo last updated on 24/Sep/19
let S_n =Σ_(k=1) ^n (((−1)^k )/(k^2 (k+1)^3 )) first 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 )) lim_(x→+∞) xF(x)=0 =a+c ⇒c=−a ⇒ F(x)=(a/x)−(a/(x+1)) +(1/x^2 ) +(d/((x+1)^2 )) +(1/((x+1)^3 )) =(a/(x^2 +x)) +(1/x^2 ) +(d/((x+1)^2 )) +(1/((x+1)^3 )) ⇒lim_(x→+∞) x^2 F(x) =a+1 +d=0 ⇒d =−a−1 ⇒ F(x)=(a/x)−(a/(x+1)) +(1/x^2 )−((a+1)/((x+1)^2 )) +(1/((x+1)^3 )) F(−2) =−(1/4) =−(a/2) +a +(1/4)−a−1−1 =−(a/2)−(7/4) ⇒ (1/4) =(a/2)+(7/4) ⇒1 =2a+7 ⇒2a=−6 ⇒a=−3 ⇒ F(x)=−(3/x) +(3/(x+1)) +(1/x^2 ) +(2/((x+1)^2 )) +(1/((x+1)^3 )) ⇒ S_n =Σ_(k=1) ^n (−1)^k F(k)=−3Σ_(k=1) ^n (((−1)^k )/k) +3Σ_(k=1) ^n (((−1)^k )/(k+1)) +Σ_(k=1) ^n (((−1)^k )/k^2 ) +2Σ_(k=1) ^n (((−1)^k )/((k+1)^2 )) +Σ_(k=1) ^n (((−1)^k )/((k+1)^3 )) Σ_(k=1) ^n (((−1)^k )/k) →−ln(2) Σ_(k=1) ^n (((−1)^k )/(k+1)) =Σ_(k=2) ^(n+1) (((−1)^(k−1) )/k) →ln(2)−1 Σ_(k=1) ^n (((−1)^k )/k^2 ) →Σ_(k=1) ^∞ (((−1)^k )/k^2 ) =(2^(1−2) −1)ξ(2)=−(1/2)(π^2 /6) =−(π^2 /(12)) Σ_(k=1) ^n (((−1)^k )/((k+1)^2 )) =Σ_(k=2) ^(n+1) (((−1)^(k−1) )/k^2 ) =−Σ_(k=1) ^(n+1) (((−1)^k )/k^2 )−1→(π^2 /(12))−1 Σ_(k=1) ^n (((−1)^k )/((k+1)^3 )) =Σ_(k=2) ^(n+1) (((−1)^(k−1) )/k^3 ) =−Σ_(k=1) ^(n+1) (((−1)^k )/k^3 ) −1 →−(2^(1−3) −1)ξ(3)−1 =(1−(1/4))ξ(3)−1 =(3/4)ξ(3)−1 ⇒ S =lim_(n→+∞) S_n =3ln(2)+3ln(2)−3−(π^2 /(12)) +(π^2 /6)−2 +(3/4)ξ(3)−1 S=6ln(2) +(π^2 /(12)) +(3/4)ξ(3) −6
$${let}\:{S}_{{n}} =\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}^{\mathrm{2}} \left({k}+\mathrm{1}\right)^{\mathrm{3}} }\:\:{first}\:{let}\:{decompose}\:{F}\left({x}\right)=\frac{\mathrm{1}}{{x}^{\mathrm{2}} \left({x}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$$${F}\left({x}\right)=\frac{{a}}{{x}}\:+\frac{{b}}{{x}^{\mathrm{2}} }\:+\frac{{c}}{{x}+\mathrm{1}}\:+\frac{{d}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }\:+\frac{{e}}{\left({x}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$$${b}={lim}_{{x}\rightarrow\mathrm{0}} \:\:{x}^{\mathrm{2}} {F}\left({x}\right)=\mathrm{1} \\ $$$${e}={lim}_{{x}\rightarrow−\mathrm{1}} \left({x}+\mathrm{1}\right)^{\mathrm{3}} {F}\left({x}\right)=\mathrm{1}\:\Rightarrow{F}\left({x}\right)=\frac{{a}}{{x}}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:+\frac{{c}}{{x}+\mathrm{1}}\:+\frac{{d}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }\:+\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$$${lim}_{{x}\rightarrow+\infty} {xF}\left({x}\right)=\mathrm{0}\:={a}+{c}\:\Rightarrow{c}=−{a}\:\Rightarrow \\ $$$${F}\left({x}\right)=\frac{{a}}{{x}}−\frac{{a}}{{x}+\mathrm{1}}\:+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:+\frac{{d}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }\:+\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$$$=\frac{{a}}{{x}^{\mathrm{2}} +{x}}\:+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:+\frac{{d}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }\:+\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)^{\mathrm{3}} }\:\Rightarrow{lim}_{{x}\rightarrow+\infty} {x}^{\mathrm{2}} {F}\left({x}\right) \\ $$$$={a}+\mathrm{1}\:+{d}=\mathrm{0}\:\Rightarrow{d}\:=−{a}−\mathrm{1}\:\Rightarrow \\ $$$${F}\left({x}\right)=\frac{{a}}{{x}}−\frac{{a}}{{x}+\mathrm{1}}\:+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }−\frac{{a}+\mathrm{1}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }\:+\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$$${F}\left(−\mathrm{2}\right)\:=−\frac{\mathrm{1}}{\mathrm{4}}\:=−\frac{{a}}{\mathrm{2}}\:+{a}\:+\frac{\mathrm{1}}{\mathrm{4}}−{a}−\mathrm{1}−\mathrm{1}\:=−\frac{{a}}{\mathrm{2}}−\frac{\mathrm{7}}{\mathrm{4}}\:\Rightarrow \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\:=\frac{{a}}{\mathrm{2}}+\frac{\mathrm{7}}{\mathrm{4}}\:\Rightarrow\mathrm{1}\:=\mathrm{2}{a}+\mathrm{7}\:\Rightarrow\mathrm{2}{a}=−\mathrm{6}\:\Rightarrow{a}=−\mathrm{3}\:\Rightarrow \\ $$$${F}\left({x}\right)=−\frac{\mathrm{3}}{{x}}\:+\frac{\mathrm{3}}{{x}+\mathrm{1}}\:+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:+\frac{\mathrm{2}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }\:+\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)^{\mathrm{3}} }\:\Rightarrow \\ $$$${S}_{{n}} =\sum_{{k}=\mathrm{1}} ^{{n}} \left(−\mathrm{1}\right)^{{k}} \:{F}\left({k}\right)=−\mathrm{3}\sum_{{k}=\mathrm{1}} ^{{n}} \frac{\left(−\mathrm{1}\right)^{{k}} }{{k}}\:+\mathrm{3}\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}+\mathrm{1}} \\ $$$$+\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}^{\mathrm{2}} }\:+\mathrm{2}\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\left(−\mathrm{1}\right)^{{k}} }{\left({k}+\mathrm{1}\right)^{\mathrm{2}} }\:+\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\left(−\mathrm{1}\right)^{{k}} }{\left({k}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$$$\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}}\:\rightarrow−{ln}\left(\mathrm{2}\right) \\ $$$$\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}+\mathrm{1}}\:=\sum_{{k}=\mathrm{2}} ^{{n}+\mathrm{1}} \:\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} }{{k}}\:\rightarrow{ln}\left(\mathrm{2}\right)−\mathrm{1} \\ $$$$\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}^{\mathrm{2}} }\:\rightarrow\sum_{{k}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}^{\mathrm{2}} }\:=\left(\mathrm{2}^{\mathrm{1}−\mathrm{2}} −\mathrm{1}\right)\xi\left(\mathrm{2}\right)=−\frac{\mathrm{1}}{\mathrm{2}}\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:=−\frac{\pi^{\mathrm{2}} }{\mathrm{12}} \\ $$$$\sum_{{k}=\mathrm{1}} ^{{n}} \frac{\left(−\mathrm{1}\right)^{{k}} }{\left({k}+\mathrm{1}\right)^{\mathrm{2}} }\:=\sum_{{k}=\mathrm{2}} ^{{n}+\mathrm{1}} \:\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} }{{k}^{\mathrm{2}} }\:=−\sum_{{k}=\mathrm{1}} ^{{n}+\mathrm{1}} \:\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}^{\mathrm{2}} }−\mathrm{1}\rightarrow\frac{\pi^{\mathrm{2}} }{\mathrm{12}}−\mathrm{1} \\ $$$$\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\left(−\mathrm{1}\right)^{{k}} }{\left({k}+\mathrm{1}\right)^{\mathrm{3}} }\:=\sum_{{k}=\mathrm{2}} ^{{n}+\mathrm{1}} \:\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} }{{k}^{\mathrm{3}} }\:=−\sum_{{k}=\mathrm{1}} ^{{n}+\mathrm{1}} \:\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}^{\mathrm{3}} }\:−\mathrm{1} \\ $$$$\rightarrow−\left(\mathrm{2}^{\mathrm{1}−\mathrm{3}} −\mathrm{1}\right)\xi\left(\mathrm{3}\right)−\mathrm{1}\:=\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}\right)\xi\left(\mathrm{3}\right)−\mathrm{1}\:=\frac{\mathrm{3}}{\mathrm{4}}\xi\left(\mathrm{3}\right)−\mathrm{1}\:\Rightarrow \\ $$$${S}\:={lim}_{{n}\rightarrow+\infty} \:{S}_{{n}} \\ $$$$=\mathrm{3}{ln}\left(\mathrm{2}\right)+\mathrm{3}{ln}\left(\mathrm{2}\right)−\mathrm{3}−\frac{\pi^{\mathrm{2}} }{\mathrm{12}}\:+\frac{\pi^{\mathrm{2}} }{\mathrm{6}}−\mathrm{2}\:+\frac{\mathrm{3}}{\mathrm{4}}\xi\left(\mathrm{3}\right)−\mathrm{1} \\ $$$${S}=\mathrm{6}{ln}\left(\mathrm{2}\right)\:+\frac{\pi^{\mathrm{2}} }{\mathrm{12}}\:+\frac{\mathrm{3}}{\mathrm{4}}\xi\left(\mathrm{3}\right)\:−\mathrm{6} \\ $$

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