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Question Number 72889 by mathmax by abdo last updated on 04/Nov/19
calculate Σ_(n=1) ^∞ (((−1)^n )/((2n+1)n^2 ))
$${calculate}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right){n}^{\mathrm{2}} } \\ $$
Commented by mathmax by abdo last updated on 04/Nov/19
let S=Σ_(n=1) ^∞ (((−1)^n )/(n^2 (2n+1))) first let decompose F(x)=(1/(x^2 (2x+1))) F(x)=(a/x)+(b/x^2 ) +(c/(2x+1)) b =lim_(x→0) x^2 F(x)=1 c=lim_(x→−(1/2)) (2x+1)F(x)=4 ⇒F(x)=(a/x)+(1/x^2 )+(4/(2x+1)) lim_(x→+∞) xF(x)=0=a+(c/2) ⇒a=−2 ⇒F(x)=−(2/x)+(1/x^2 )+(4/(2x+1)) ⇒ S =Σ_(n=1) ^∞ (−1)^n F(n) =−2Σ_(n=1) ^∞ (((−1)^n )/n) +Σ_(n=1) ^∞ (((−1)^n )/n^2 ) +4Σ_(n=1) ^∞ (((−1)^n )/(2n+1)) Σ_(n=1) ^∞ (((−1)^n )/n) =−ln(2) Σ_(n=1) ^∞ (((−1)^n )/n^2 ) =(2^(1−2) −1)ξ(2) =−(π^2 /(12)) Σ_(n=1) ^∞ (((−1)^n )/(2n+1)) =Σ_(n=0) ^∞ (((−1)^n )/(2n+1))−1 =(π/4)−1 ⇒ S =−2(−ln2)−(π^2 /(12)) +4((π/4)−1) =2ln(2)−(π^2 /(12)) +π −4
$${let}\:{S}=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} \left(\mathrm{2}{n}+\mathrm{1}\right)}\:\:{first}\:{let}\:{decompose}\:{F}\left({x}\right)=\frac{\mathrm{1}}{{x}^{\mathrm{2}} \left(\mathrm{2}{x}+\mathrm{1}\right)} \\ $$$${F}\left({x}\right)=\frac{{a}}{{x}}+\frac{{b}}{{x}^{\mathrm{2}} }\:+\frac{{c}}{\mathrm{2}{x}+\mathrm{1}} \\ $$$${b}\:={lim}_{{x}\rightarrow\mathrm{0}} {x}^{\mathrm{2}} {F}\left({x}\right)=\mathrm{1} \\ $$$${c}={lim}_{{x}\rightarrow−\frac{\mathrm{1}}{\mathrm{2}}} \:\:\left(\mathrm{2}{x}+\mathrm{1}\right){F}\left({x}\right)=\mathrm{4}\:\Rightarrow{F}\left({x}\right)=\frac{{a}}{{x}}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }+\frac{\mathrm{4}}{\mathrm{2}{x}+\mathrm{1}} \\ $$$${lim}_{{x}\rightarrow+\infty} {xF}\left({x}\right)=\mathrm{0}={a}+\frac{{c}}{\mathrm{2}}\:\Rightarrow{a}=−\mathrm{2}\:\Rightarrow{F}\left({x}\right)=−\frac{\mathrm{2}}{{x}}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }+\frac{\mathrm{4}}{\mathrm{2}{x}+\mathrm{1}}\:\Rightarrow \\ $$$${S}\:=\sum_{{n}=\mathrm{1}} ^{\infty} \left(−\mathrm{1}\right)^{{n}} {F}\left({n}\right)\:=−\mathrm{2}\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}}\:+\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} }\:+\mathrm{4}\sum_{{n}=\mathrm{1}} ^{\infty} \frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}{n}+\mathrm{1}} \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}}\:=−{ln}\left(\mathrm{2}\right) \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} }\:=\left(\mathrm{2}^{\mathrm{1}−\mathrm{2}} −\mathrm{1}\right)\xi\left(\mathrm{2}\right)\:=−\frac{\pi^{\mathrm{2}} }{\mathrm{12}} \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}{n}+\mathrm{1}}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}{n}+\mathrm{1}}−\mathrm{1}\:=\frac{\pi}{\mathrm{4}}−\mathrm{1}\:\Rightarrow \\ $$$${S}\:=−\mathrm{2}\left(−{ln}\mathrm{2}\right)−\frac{\pi^{\mathrm{2}} }{\mathrm{12}}\:+\mathrm{4}\left(\frac{\pi}{\mathrm{4}}−\mathrm{1}\right)\:=\mathrm{2}{ln}\left(\mathrm{2}\right)−\frac{\pi^{\mathrm{2}} }{\mathrm{12}}\:+\pi\:−\mathrm{4} \\ $$
Answered by mind is power last updated on 04/Nov/19
(1/((2n+1)n^2 ))=(1/n^2 )+(c/n)+(4/(2n+1))⇒(1/n^2 )+((−2)/n^ )+(4/(2n+1)) Σ_(n≥1) (((−1)^n )/n^2 )+Σ_(n≥1) (((−1)^(n+1) )/n)+4Σ(((−1)^n )/(2n+1)) (1/(1+x^2 ))=Σ_(n≥0) (−x^2 )^n arctan(x)=Σ(−1)^n .(x^(2n+1) /(2n+1)) x=1⇒Σ(((−1)^n )/(2n+1))=(π/4)−1 Σ_(n≥1) (((−1)^(n+1) )/n)→ln(2) Σ_(n≥1) (((−1)^n )/n^2 )=(1/4).ζ(2)−(3/4)ζ(2)=−((ζ(2))/2)=−(π^2 /(12)) ⇒Σ_(n≥1) (((−1)^n )/((2n+1)n^2 ))=−(π^2 /(12))+2ln(2)+π−4
$$\frac{\mathrm{1}}{\left(\mathrm{2n}+\mathrm{1}\right)\mathrm{n}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} }+\frac{\mathrm{c}}{\mathrm{n}}+\frac{\mathrm{4}}{\mathrm{2n}+\mathrm{1}}\Rightarrow\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} }+\frac{−\mathrm{2}}{\mathrm{n}^{} }+\frac{\mathrm{4}}{\mathrm{2n}+\mathrm{1}} \\ $$$$\underset{\mathrm{n}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{n}^{\mathrm{2}} }+\underset{\mathrm{n}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{\mathrm{n}+\mathrm{1}} }{\mathrm{n}}+\mathrm{4}\Sigma\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{2n}+\mathrm{1}} \\ $$$$\frac{\mathrm{1}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }=\underset{\mathrm{n}\geqslant\mathrm{0}} {\sum}\left(−\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{n}} \\ $$$$\mathrm{arctan}\left(\mathrm{x}\right)=\Sigma\left(−\mathrm{1}\right)^{\mathrm{n}} .\frac{\mathrm{x}^{\mathrm{2n}+\mathrm{1}} }{\mathrm{2n}+\mathrm{1}} \\ $$$$\mathrm{x}=\mathrm{1}\Rightarrow\Sigma\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{2n}+\mathrm{1}}=\frac{\pi}{\mathrm{4}}−\mathrm{1} \\ $$$$\underset{\mathrm{n}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{\mathrm{n}+\mathrm{1}} }{\mathrm{n}}\rightarrow\mathrm{ln}\left(\mathrm{2}\right) \\ $$$$\underset{\mathrm{n}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{n}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{4}}.\zeta\left(\mathrm{2}\right)−\frac{\mathrm{3}}{\mathrm{4}}\zeta\left(\mathrm{2}\right)=−\frac{\zeta\left(\mathrm{2}\right)}{\mathrm{2}}=−\frac{\pi^{\mathrm{2}} }{\mathrm{12}} \\ $$$$\Rightarrow\underset{\mathrm{n}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\left(\mathrm{2n}+\mathrm{1}\right)\mathrm{n}^{\mathrm{2}} }=−\frac{\pi^{\mathrm{2}} }{\mathrm{12}}+\mathrm{2ln}\left(\mathrm{2}\right)+\pi−\mathrm{4} \\ $$
Commented by turbo msup by abdo last updated on 04/Nov/19
thsnk you sir.
$${thsnk}\:{you}\:{sir}. \\ $$
Commented by mind is power last updated on 04/Nov/19
y,re welcom
$$\mathrm{y},\mathrm{re}\:\mathrm{welcom} \\ $$

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