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Question Number 181747 by SANOGO last updated on 29/Nov/22

calcul la somme Σ_(n=2) ^(+oo) (x/((n^2 −1)^2 ))

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

Answered by Frix last updated on 01/Dec/22

Σ_(n=2) ^∞ (x/((n^2 −1)^2 )) =xΣ_(n=2) ^∞ (1/((n^2 −1))) = =xΣ_(n=2) ^∞ ((1/(4(n−1)^2 ))+(1/(4(n+1)^2 ))−(1/(2(n^2 −1)))) = =x((π^2 /(24))+(π^2 /(24))−(5/(16))−lim_(k→∞) (((k−1)(3k+2))/(8k(k+1))))= =x((π^2 /(12))−((11)/(16)))

$$\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\:\frac{{x}}{\left({n}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }\:={x}\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{\left({n}^{\mathrm{2}} −\mathrm{1}\right)}\:= \\ $$$$={x}\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\:\left(\frac{\mathrm{1}}{\mathrm{4}\left({n}−\mathrm{1}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{4}\left({n}+\mathrm{1}\right)^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{2}\left({n}^{\mathrm{2}} −\mathrm{1}\right)}\right)\:= \\ $$$$={x}\left(\frac{\pi^{\mathrm{2}} }{\mathrm{24}}+\frac{\pi^{\mathrm{2}} }{\mathrm{24}}−\frac{\mathrm{5}}{\mathrm{16}}−\underset{{k}\rightarrow\infty} {\mathrm{lim}}\frac{\left({k}−\mathrm{1}\right)\left(\mathrm{3}{k}+\mathrm{2}\right)}{\mathrm{8}{k}\left({k}+\mathrm{1}\right)}\right)= \\ $$$$={x}\left(\frac{\pi^{\mathrm{2}} }{\mathrm{12}}−\frac{\mathrm{11}}{\mathrm{16}}\right) \\ $$

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