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Question Number 86786 by M±th+et£s last updated on 31/Mar/20

$$showthat$$$$\underset{n=2}{\sum ^{\mathrm{\infty }}}\frac{1}{n^2{(1-n^2)}^2}=0.2999$$

Commented by Prithwish Sen 1 last updated on 31/Mar/20

$$\mathrm{I}\mathrm{got}0.0299$$

Commented by Prithwish Sen 1 last updated on 31/Mar/20

$$\mathrm{the}\mathrm{series}\mathrm{is}$$$$\frac{3}{2}\underset{1}{\sum ^{\mathrm{\infty }}}\frac{1}{{\mathrm{n}}^2}-\frac{39}{16}=\frac{{\mathit{\pi }}^2}{4}-\frac{39}{16}=0.0299011$$$$\mathbf{please}\mathbf{check}.$$

Commented by mathmax by abdo last updated on 31/Mar/20

$$letdecomposeF\left(x\right)=\frac{1}{x^2{(x^2-1)}^2}=\frac{1}{x^2{(x-1)}^2{(x+1)}^2}$$$$F\left(x\right)=\frac{a}{x}+\frac{b}{x^2}+\frac{c}{x-1}+\frac{d}{{(x-1)}^2}+\frac{e}{x+1}+\frac{f}{{(x+1)}^2}$$$$F(-x)=F\left(x\right)\Rightarrow -\frac{a}{x}+\frac{b}{x^2}-\frac{c}{x+1}+\frac{d}{{(x+1)}^2}-\frac{e}{x-1}+\frac{f}{{(x-1)}^2}$$$$=F\left(x\right)\Rightarrow a=0ande=-candf=d\Rightarrow$$$$f\left(x\right)=\frac{b}{x^2}+\frac{c}{x-1}+\frac{d}{{(x-1)}^2}-\frac{c}{x+1}+\frac{d}{{(x+1)}^2}$$$$b=1andd=\frac{1}{4}\Rightarrow F\left(x\right)=\frac{1}{x^2}+\frac{c}{x-1}+\frac{1}{4{(x-1)}^2}-\frac{c}{x+1}+\frac{1}{4{(x+1)}^2}$$$$F\left(2\right)=\frac{1}{36}=\frac{1}{4}+c+\frac{1}{4}-\frac{c}{3}+\frac{1}{36}\Rightarrow \frac{1}{2}+\frac{2}{3}c=0\Rightarrow 3+4c=0\Rightarrow$$$$c=-\frac{3}{4}\Rightarrow F\left(x\right)=\frac{1}{x^2}-\frac{3}{4(x-1)}+\frac{1}{4{(x-1)}^2}+\frac{3}{4(x+1)}+\frac{1}{4{(x+1)}^2}$$$$S_n=\sum _{k=2}^n\frac{1}{k^2{(k^2-1)}^2}\Rightarrow$$$$S_n=\sum _{k=2}^n\frac{1}{k^2}-\frac{3}{4}\sum _{k=2}^n\frac{1}{k-1}+\frac{1}{4}\sum _{k=2}^n\frac{1}{{(k-1)}^2}+\frac{3}{4}\sum _{k=2}^n\frac{1}{k+1}$$$$+\frac{1}{4}\sum _{k=2}^n\frac{1}{{(k+1)}^2}$$$$\sum _{k.=2}^n\frac{1}{k^2}={\xi }_n\left(2\right)-1\to \frac{{\pi }^2}{6}-1$$$$\sum _{k=2}^n\frac{1}{k-1}=\sum _{k=1}^{n-1}\frac{1}{k}=H_{n-1}=ln(n-1)+\gamma +o\left(\frac{1}{n-1}\right)$$$$\sum _{k=2}^n\frac{1}{{(k-1)}^2}=\sum _{k=1}^{n-1}\frac{1}{k^2}={\xi }_{n-1}\left(2\right)\to \frac{{\pi }^2}{6}$$$$\sum _{k=2}^n\frac{1}{k+1}=\sum _{k=3}^{n+1}\frac{1}{k}=H_{n+1}-\frac{3}{2}=ln(n+1)+\gamma +o\left(\frac{1}{n+1}\right)-\frac{3}{2}$$$$\sum _{k=2}^n\frac{1}{{(k+1)}^2}=\sum _{k=3}^{n+1}\frac{1}{k^2}={\xi }_{n+1}\left(2\right)-\frac{5}{4}\to \frac{{\pi }^2}{6}-\frac{5}{4}\Rightarrow$$$$S_n={\xi }_n\left(2\right)-1-\frac{3}{4}{H}_{n-1}+\frac{1}{4}{\xi }_{n-1}\left(2\right)+\frac{3}{4}{H}_{n+1}-\frac{9}{8}+\frac{1}{4}{\xi }_{n+1}\left(2\right)-\frac{5}{16}$$$$\Rightarrow {lim}_{n\to +\mathrm{\infty }}{S}_n=\frac{{\pi }^2}{6}-1+\frac{1}{4}\frac{{\pi }^2}{6}-\frac{9}{8}+\frac{1}{4}\frac{{\pi }^2}{6}-\frac{5}{16}$$$$=\frac{{\pi }^2}{6}+\frac{{\pi }^2}{12}-\frac{16+18+5}{16}=\frac{{\pi }^2}{4}-\frac{39}{16}$$

Commented by M±th+et£s last updated on 31/Mar/20

$$yessiritstypo$$

Commented by M±th+et£s last updated on 31/Mar/20

$$thankyou$$

Commented by mathmax by abdo last updated on 01/Apr/20

$$youare[welcome$$

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