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Question Number 183827 by SANOGO last updated on 30/Dec/22

calcul: S=Σ_(n=2 ) ^(+oo) (n/((n^2 −1)^2 ))

$${calcul}:\:\:{S}=\underset{{n}=\mathrm{2}\:\:} {\overset{+{oo}} {\sum}}\frac{{n}}{\left({n}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} } \\ $$

Answered by Ar Brandon last updated on 30/Dec/22

S=Σ_(n=2) ^∞ (n/((n^2 −1)^2 ))=Σ_(n=2) ^∞ (n/(((n−1)(n+1))^2 )) =(1/4)Σ_(n=2) ^∞ ((n((n+1)−(n−1))^2 )/((n−1)^2 (n+1)^2 )) =(1/4)Σ_(n=2) ^∞ ((n/((n−1)^2 ))−((2n)/((n−1)(n+1)))+(n/((n+1)^2 ))) =(1/4)Σ_(n=2) ^∞ ((1/(n−1))+(1/((n−1)^2 ))−(1/(n−1))−(1/(n+1))+(1/(n+1))−(1/((n+1)^2 ))) =(1/4)Σ_(n=2) ((1/((n−1)^2 ))−(1/((n+1)^2 )))=(1/4)(Σ_(n=1) ^∞ (1/n^2 )−(Σ_(n=1) ^∞ (1/n^2 )−1−(1/2^2 ))) =(1/4)(1+(1/4))=(5/(16))

$${S}=\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{{n}}{\left({n}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }=\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{{n}}{\left(\left({n}−\mathrm{1}\right)\left({n}+\mathrm{1}\right)\right)^{\mathrm{2}} } \\ $$$$\:\:\:=\frac{\mathrm{1}}{\mathrm{4}}\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{{n}\left(\left({n}+\mathrm{1}\right)−\left({n}−\mathrm{1}\right)\right)^{\mathrm{2}} }{\left({n}−\mathrm{1}\right)^{\mathrm{2}} \left({n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\:\:\:=\frac{\mathrm{1}}{\mathrm{4}}\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\left(\frac{{n}}{\left({n}−\mathrm{1}\right)^{\mathrm{2}} }−\frac{\mathrm{2}{n}}{\left({n}−\mathrm{1}\right)\left({n}+\mathrm{1}\right)}+\frac{{n}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\right) \\ $$$$\:\:\:=\frac{\mathrm{1}}{\mathrm{4}}\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{{n}−\mathrm{1}}+\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right)^{\mathrm{2}} }−\frac{\mathrm{1}}{{n}−\mathrm{1}}−\frac{\mathrm{1}}{{n}+\mathrm{1}}+\frac{\mathrm{1}}{{n}+\mathrm{1}}−\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\right) \\ $$$$\:\:\:=\frac{\mathrm{1}}{\mathrm{4}}\underset{{n}=\mathrm{2}} {\sum}\left(\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right)^{\mathrm{2}} }−\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\right)=\frac{\mathrm{1}}{\mathrm{4}}\left(\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} }−\left(\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} }−\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }\right)\right) \\ $$$$\:\:\:=\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}}\right)=\frac{\mathrm{5}}{\mathrm{16}} \\ $$

Commented by MJS_new last updated on 30/Dec/22

nice!

$$\mathrm{nice}! \\ $$

Commented by Ar Brandon last updated on 30/Dec/22

Thanks master��

Commented by SANOGO last updated on 30/Dec/22

merci

$${merci} \\ $$

Commented by MJS_new last updated on 31/Dec/22

I'm still a beginner, maybe with extra years of experience ��

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