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

$$calcul:S=\underset{n=2}{\sum ^{+oo}}\frac{n}{{(n^2-1)}^2}$$

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

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

Commented by MJS_new last updated on 30/Dec/22

$$\mathrm{nice}!$$

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

Thanks master��

Commented by SANOGO last updated on 30/Dec/22

$$merci$$

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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