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Question Number 33713 by abdo imad last updated on 22/Apr/18

$$findtbevalueof\sum _{n=2}^{\mathrm{\infty }}\frac{n^2-n+1}{{(n-1)}^2{(n+1)}^2}.$$

Commented by prof Abdo imad last updated on 26/Apr/18

$$letput{S}_n=\sum _{k=2}^n\frac{n^2-n+1}{{(n-1)}^2{(n+1)}^2}let?drcompose$$$$F\left(x\right)=\frac{x^2-x+1}{{(x-1)}^2{(x+1)}^2}$$$$F\left(x\right)=\frac{a}{x-1}+\frac{b}{{(x-1)}^2}+\frac{b}{x+1}+\frac{c}{{(x+1)}^2}$$$$b={lim}_{x\to 1}{(x-1)}^{2}F\left(x\right)=\frac{1}{4}$$$$c={lim}_{x\to -1}{(x+1)}^{2}F\left(x\right)=\frac{3}{4}\Rightarrow$$$$F\left(x\right)=\frac{a}{x-1}+\frac{1}{4{(x-1)}^2}+\frac{b}{x+1}+\frac{3}{4{(x+1)}^2}$$$$F\left(0\right)=1=-a+\frac{1}{4}+b+\frac{3}{4}=1+b-a\Rightarrow b=a\Rightarrow$$$$F\left(x\right)=\frac{a}{x-1}+\frac{1}{4{(x-1)}^2}+\frac{a}{x+1}+\frac{3}{4{(x+1)}^2}$$$$=\frac{2ax}{x^2-1}+\frac{1}{4{(x-1)}^2}+\frac{3}{4{(x+1)}^2}$$$${lim}_{x\to +\mathrm{\infty }}xF\left(x\right)=0=2a\Rightarrow a=0\Rightarrow$$$$F\left(x\right)=\frac{1}{4{(x-1)}^2}+\frac{3}{4{(x+1)}^2}\Rightarrow$$$$S_n=\frac{1}{4}\sum _{k=2}^n\frac{1}{{(k-1)}^2}+\frac{3}{4}\sum _{k=2}^n\frac{1}{{(k+1)}^2}but$$$$\sum _{k=2}^n\frac{1}{{(k-1)}^2}=\sum _{k=1}^{n-1}\frac{1}{k^2}={\xi }_{n-1}\left(2\right)\to \xi \left(2\right)=\frac{{\pi }^2}{6}$$$$\sum _{k=2}^n\frac{1}{{(k+1)}^2}=\sum _{k=3}^{n+1}\frac{1}{k^2}=\sum _{k=1}^{n+1}\frac{1}{k^2}-1-\frac{1}{4}$$$$={\xi }_{n+1}\left(2\right)-\frac{3}{4}\to \xi \left(2\right)-\frac{3}{4}=\frac{{\pi }^2}{6}-\frac{3}{4}$$$${lim}_{n\to +\mathrm{\infty }}{S}_n=\frac{1}{4}\frac{{\pi }^2}{6}+\frac{3}{4}(\frac{{\pi }^2}{6}-\frac{3}{4})$$$$=\frac{{\pi }^2}{24}+\frac{3{\pi }^2}{24}-\frac{9}{16}=\frac{{\pi }^2}{6}-\frac{9}{16}.$$$$$$

Commented by sma3l2996 last updated on 26/Apr/18

$$Ithinkyouhaveanerroronlines14and15$$$$-1-\frac{1}{4}=\frac{-5}{4}$$

Commented by prof Abdo imad last updated on 26/Apr/18

$$yesyesyouarerightiwillcorrecttheanswer$$$$thankyousomuchsirsma3l....$$

Commented by prof Abdo imad last updated on 26/Apr/18

$$errorfromline15\sum _{k=2}^n\frac{1}{{(k+1)}^2}={\xi }_{n+1}\left(2\right)-\frac{5}{4}$$$${\to }_{n\to +\mathrm{\infty }}\xi \left(2\right)-\frac{5}{4}=\frac{{\pi }^2}{6}-\frac{5}{4}\Rightarrow$$$${lim}_{n\to +\mathrm{\infty }}{S}_n=\frac{1}{4}\frac{{\pi }^2}{6}+\frac{3}{4}(\frac{{\pi }^2}{6}-\frac{5}{4})$$$$=\frac{{\pi }^2}{24}+\frac{3{\pi }^2}{24}-\frac{15}{16}=\frac{{\pi }^2}{6}-\frac{15}{16}$$$$★lim{S}_n=\frac{{\pi }^2}{6}-\frac{15}{16}★$$

Answered by sma3l2996 last updated on 23/Apr/18

$$A_n=\underset{n=2}{\sum ^{\mathrm{\infty }}}\frac{n^2-n+1}{{(n-1)}^2{(n+1)}^2}=\underset{n=2}{\sum ^{\mathrm{\infty }}}(\frac{a}{n-1}+\frac{b}{{(n-1)}^2}+\frac{c}{n+1}+\frac{d}{{(n+1)}^2})$$$$witha=c=0;b=\frac{1}{4};d=\frac{3}{4}$$$$\underset{n=2}{\sum ^{\mathrm{\infty }}}\frac{n^2-n+1}{{(n-1)}^2{(n+1)}^2}=\frac{1}{4}\underset{n=2}{\sum ^{\mathrm{\infty }}}(\frac{1}{{(n-1)}^2}+\frac{3}{{(n+1)}^2})$$$$=\frac{1}{4}\underset{n=2}{\sum ^{\mathrm{\infty }}}\frac{1}{{(n-1)}^2}+\frac{3}{4}\underset{n=2}{\sum ^{\mathrm{\infty }}}\frac{1}{{(n+1)}^2}$$$$letm=n-1andk=n+1$$$$so\underset{n=2}{\sum ^{\mathrm{\infty }}}\frac{n^2-n+1}{{(n-1)}^2{(n+1)}^2}=\frac{1}{4}\underset{m=1}{\sum ^{\mathrm{\infty }}}\frac{1}{m^2}+\frac{3}{4}\underset{k=3}{\sum ^{\mathrm{\infty }}}\frac{1}{k^2}$$$$=\frac{1}{4}\xi \left(2\right)+\frac{3}{4}(\xi \left(2\right)-(\frac{1}{1^2}+\frac{1}{2^2}))=\xi \left(2\right)-\frac{5}{4}with\left(\xi \left(x\right)iszetafunction\right)$$$$=\frac{{\pi }^2}{6}-\frac{5}{4}$$

Commented by prof Abdo imad last updated on 26/Apr/18

$$youhavecommitedaerrorofcalculussirsma3l..$$

Commented by sma3l2996 last updated on 26/Apr/18

$$\frac{1}{4}\xi \left(2\right)+\frac{3}{4}(\xi \left(2\right)-\frac{5}{4})=\xi \left(2\right)-\frac{15}{16}$$$$=\frac{{\pi }^2}{6}-\frac{15}{16}$$

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