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Question Number 95216 by mathmax by abdo last updated on 24/May/20

$$\mathrm{calculste}\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{\mathrm{n}}{{(2\mathrm{n}+1)}^2(\mathrm{n}+3)}$$

Answered by abdomathmax last updated on 25/May/20

$$\mathrm{let}{\mathrm{S}}_{\mathrm{n}}=\sum _{\mathrm{k}=0}^{\mathrm{n}}\frac{\mathrm{k}}{(\mathrm{k}+3){(2\mathrm{k}+1)}^2}$$$$\mathrm{let}\mathrm{decompose}\mathrm{F}\left(\mathrm{x}\right)=\frac{\mathrm{x}}{(\mathrm{x}+3){(2\mathrm{x}+1)}^2}$$$$\mathrm{F}\left(\mathrm{x}\right)=\frac{\mathrm{a}}{\mathrm{x}+3}+\frac{\mathrm{b}}{(2\mathrm{x}+1)}+\frac{\mathrm{c}}{{(2\mathrm{x}+1)}^2}$$$$\mathrm{a}=\frac{-3}{{(-5)}^2}=-\frac{3}{25}$$$$\mathrm{c}=\frac{-1}{2(3-\frac{1}{2})}=-\frac{1}{5}$$$${\lim }_{\mathrm{x}\to +\mathrm{\infty }}\mathrm{xF}\left(\mathrm{x}\right)=0=\mathrm{a}+\frac{\mathrm{b}}{2}\Rightarrow \frac{\mathrm{b}}{2}=\frac{3}{25}\Rightarrow \mathrm{b}=\frac{6}{25}\Rightarrow$$$${\mathrm{S}}_{\mathrm{n}}=-\frac{3}{25}\sum _{\mathrm{k}=0}^{\mathrm{n}}\frac{1}{\mathrm{k}+3}+\frac{6}{25}\sum _{\mathrm{k}=0}^{\mathrm{n}}\frac{1}{2\mathrm{k}+1}-\frac{1}{5}\sum _{\mathrm{k}=0}^{\mathrm{n}}\frac{1}{{(2\mathrm{k}+1)}^2}$$$$\mathrm{we}\mathrm{have}\sum _{\mathrm{k}=0}^{\mathrm{n}}\frac{1}{\mathrm{k}+3}=\sum _{\mathrm{k}=3}^{\mathrm{n}+3}\frac{1}{\mathrm{k}}={\mathrm{H}}_{\mathrm{n}+3}-\frac{3}{2}$$$$\sum _{\mathrm{k}=0}^{\mathrm{n}}\frac{1}{2\mathrm{k}+1}={\mathrm{H}}_{2\mathrm{n}+1}-\frac{1}{2}{\mathrm{H}}_{\mathrm{n}}$$$$\Rightarrow {\mathrm{S}}_{\mathrm{n}}=-\frac{3}{25}({\mathrm{H}}_{\mathrm{n}+3}-\frac{3}{2})+\frac{6}{25}({\mathrm{H}}_{2\mathrm{n}+1}-\frac{1}{2}{\mathrm{H}}_{\mathrm{n}})$$$$-\frac{1}{5}\sum _{\mathrm{k}=0}^{\mathrm{n}}\frac{1}{{(2\mathrm{k}+1)}^2}$$$$=\frac{3}{25}({\mathrm{H}}_{2\mathrm{n}+1}-{\mathrm{H}}_{\mathrm{n}+3})+\frac{3}{25}({\mathrm{H}}_{2\mathrm{n}+1}-{\mathrm{H}}_{\mathrm{n}})-\frac{1}{5}\sum _{\mathrm{k}=0}^{\mathrm{n}}\frac{1}{{(2\mathrm{k}+1)}^2}+\frac{9}{50}$$$$\Rightarrow \lim {\mathrm{S}}_{\mathrm{n}}=\frac{3}{25}\ln \left(2\right)+\frac{3}{25}\ln \left(2\right)-\frac{1}{5}\sum _{\mathrm{k}=0}^{\mathrm{\infty }}\frac{1}{{(2\mathrm{k}+1)}^2}$$$$=\frac{6}{25}\ln \left(2\right)+\frac{9}{50}-\frac{1}{5}\sum _{\mathrm{k}=0}^{\mathrm{\infty }}\frac{1}{{(2\mathrm{k}+1)}^2}$$$$\frac{{\pi }^2}{6}=\sum _{\mathrm{k}=1}^{\mathrm{\infty }}\frac{1}{{\mathrm{k}}^2}=\sum _{\mathrm{k}=1}^{\mathrm{\infty }}\frac{1}{{4\mathrm{k}}^2}+\sum _{\mathrm{k}=0}^{\mathrm{\infty }}\frac{1}{{(2\mathrm{k}+1)}^2}\Rightarrow$$$$\sum _{\mathrm{k}=0}^{\mathrm{\infty }}\frac{1}{{(2\mathrm{k}+1)}^2}=\frac{{\pi }^2}{6}-\frac{1}{4}\times \frac{{\pi }^2}{6}=\frac{3}{4}\times \frac{{\pi }^2}{6}=\frac{{\pi }^2}{8}\Rightarrow$$$$\mathrm{S}=\frac{6}{25}\ln \left(2\right)+\frac{9}{50}-\frac{{\pi }^2}{40}$$

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