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Question Number 28600 by A1B1C1D1 last updated on 27/Jan/18

Commented by abdo imad last updated on 27/Jan/18

$$fromwherecome\frac{3}{8}sir?thereisnomistakeinmyresponse$$$$andifyoufindsomethinginformme...$$

Commented by abdo imad last updated on 27/Jan/18

$$=\sum _{k=1}^{\propto }\frac{k}{{(k^2+2)}^2-4k^2}=\sum _{k=1}^{\propto }\frac{k}{(k^2+2-2k)(k^2+2+2k)}$$$$=\frac{1}{4}\sum _{k=1}^{\propto }(\frac{1}{k^2-2k+2}-\frac{1}{k^2+2k+2})$$$$letput{S}_n=\sum _{k=1}^n(\frac{1}{k^2-2k+2}-\frac{1}{k^2+2k+2})and$$$$v_k=\frac{1}{k^2-2k+2}wehave{v}_{k+1}=\frac{1}{{(k+1)}^2-2(k+1)+2}$$$$=\frac{1}{k^2+2k+1-2k-2+2}=\frac{1}{k^2+1}wehave{v}_k=\frac{1}{{(k-1)}^2+1}$$$$S_n=\sum _{k=1}^n(v_k-v_{k+1})=v_1-v_2+v_2-v_3+....v_n-v_{n+1}$$$$=v_1-v_{n+1}=1-\frac{1}{n^2+1}lim{S}_n=1so$$$${lim}_{n\to +\mathrm{\infty }}\sum _{k=1}^n\frac{1}{k^4+4}=\frac{1}{4}$$$$$$

Commented by A1B1C1D1 last updated on 27/Jan/18

$$\mathrm{The}\mathrm{correct}\mathrm{is}\frac{3}{8}{}^{\prime }{-}^{\prime }$$

Commented by A1B1C1D1 last updated on 28/Jan/18

$$\mathrm{This}\mathrm{was}\mathrm{the}\mathrm{feedback}\mathrm{i}\mathrm{was}\mathrm{given}...$$$$\mathrm{Thank}\mathrm{you}\mathrm{anyway}.$$

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