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Question Number 33591 by abdo imad last updated on 19/Apr/18

$$findthevalueof\sum _{n=1}^{\mathrm{\infty }}\frac{2}{n^3+3n^2+2n}.$$

Commented by abdo imad last updated on 20/Apr/18

$$letput{S}_n=\sum _{k=1}^n\frac{2}{k^3+3k^2+2k}wehave$$$$k^3+3k^2+2k=k(k^2+3k+2)=k(k^2+k+2k+2)$$$$=k(k(k+1)+2(k+1))=k(k+1)(k+2)letdecompose$$$$F\left(x\right)=\frac{2}{x(x+1)(x+2)}=\frac{a}{x}+\frac{b}{x+1}+\frac{c}{x+2}$$$$a={lim}_{x\to 0}xF\left(x\right)=1$$$$b={lim}_{x\to -1}(x+1)F\left(x\right)=\frac{2}{(-1).1}=-2$$$$c={lim}_{x\to -2}(x+2)F\left(x\right)=\frac{2}{(-2)(-1)}=1\Rightarrow$$$$F\left(x\right)=\frac{1}{x}-\frac{2}{x+1}+\frac{1}{x+2}and$$$$S_n=\sum _{k=1}^n(\frac{1}{k}-\frac{2}{k+1}+\frac{1}{k+2})=\sum _{k=1}^n\frac{1}{k}-2\sum _{k=1}^n\frac{1}{k+1}+\sum _{k=1}^n\frac{1}{k+2}$$$$\sum _{k=1}^n\frac{1}{k}=H_n$$$$\sum _{k=1}^n\frac{1}{k+1}=\sum _{k=2}^{n+1}\frac{1}{k}=H_{n+1}-1$$$$\sum _{k=1}^n\frac{1}{k+2}=\sum _{k=3}^{n+2}\frac{1}{k}=H_{n+2}-\frac{3}{2}\Rightarrow$$$$S_n=H_n-2(H_{n+1}-1)+H_{n+2}-\frac{3}{2}$$$$S_n=H_n+H_{n+2}-2H_{n+1}+2-\frac{3}{2}=H_n+H_{n+2}-2H_{n+1}+\frac{1}{2}$$$$but{H}_n=ln\left(n\right)+\gamma +o\left(\frac{1}{n}\right)$$$$H_{n+2}=ln(n+2)+\gamma +o\left(\frac{1}{n}\right)$$$$H_{n+1}=ln(n+1)+\gamma +o\left(\frac{1}{n}\right)\Rightarrow$$$$S_n=ln\left(n(n+2)\right)+2\gamma -2ln(n+1)-2\gamma +o\left(\frac{1}{n}\right)+\frac{1}{2}$$$$=ln\left(\frac{n^2+2n}{{(n+1)}^2}\right)+\frac{1}{2}\Rightarrow {lim}_{n\to +\mathrm{\infty }}{S}_n=\frac{1}{2}\Rightarrow$$$$★\sum _{n=1}^{\mathrm{\infty }}\frac{2}{n^3+3n^2+2n}=\frac{1}{2}★$$

Answered by sma3l2996 last updated on 19/Apr/18

$$$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{2}{n^3+3n^2+2n}=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{2}{n(n+1)(n+2)}$$$$\frac{1}{n(n+1)(n+2)}=\frac{a}{n}+\frac{b}{n+1}+\frac{c}{n+2}$$$$a=\frac{1}{2};b=-1;c=\frac{1}{2}$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{2}{n(n+1)(n+2)}=\underset{n=1}{\sum ^{\mathrm{\infty }}}(\frac{1}{n}-\frac{2}{n+1}+\frac{1}{n+2})$$$$=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n}-\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{2}{n+1}+\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n+2}$$$$letl=n+1andk=n+2$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{2}{n(n+1)(n+2)}=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n}-\underset{l=2}{\sum ^{\mathrm{\infty }}}\frac{2}{l}+\underset{k=3}{\sum ^{\mathrm{\infty }}}\frac{1}{k}$$$$=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n}-(\underset{l=1}{\sum ^{\mathrm{\infty }}}\frac{2}{l}-\frac{2}{1})+(\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{1}{k}-\frac{1}{1}-\frac{1}{2})$$$$=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n}-\underset{l=1}{\sum ^{\mathrm{\infty }}}\frac{2}{l}+\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{1}{k}+2-1-\frac{1}{2}$$$$=\underset{n=1}{\sum ^{\mathrm{\infty }}}(\frac{1}{n}-\frac{2}{n}+\frac{1}{n})+\frac{1}{2}=\frac{1}{2}$$

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