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Question Number 144831 by mathdanisur last updated on 29/Jun/21

$$\frac{3}{1\cdot 2\cdot 3}+\frac{5}{2\cdot 3\cdot 4}+\frac{7}{3\cdot 4\cdot 5}+\frac{9}{4\cdot 5\cdot 6}+...\mathrm{\infty }=?$$

Answered by Dwaipayan Shikari last updated on 29/Jun/21

$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{2n+1}{n(n+1)(n+2)}=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n(n+2)}+\frac{1}{(n+1)(n+2)}$$$$=\frac{1}{2}\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n}-\frac{1}{n+2}+\underset{n=2}{\sum ^{\mathrm{\infty }}}\frac{1}{n}-\frac{1}{n+1}$$$$=\frac{1}{2}(1+\frac{1}{2})+\frac{1}{2}=\frac{5}{4}$$

Commented by mathdanisur last updated on 29/Jun/21

$$alotcoolthanksSir$$

Answered by mathmax by abdo last updated on 30/Jun/21

$$\mathrm{S}=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{2\mathrm{n}+1}{\mathrm{n}(\mathrm{n}+1)(\mathrm{n}+2)}\mathrm{let}\mathrm{decompose}$$$$\mathrm{F}\left(\mathrm{x}\right)=\frac{2\mathrm{x}+1}{\mathrm{x}(\mathrm{x}+1)(\mathrm{x}+2)}\Rightarrow \mathrm{F}\left(\mathrm{x}\right)=\frac{\mathrm{a}}{\mathrm{x}}+\frac{\mathrm{b}}{\mathrm{x}+1}+\frac{\mathrm{c}}{\mathrm{x}+2}$$$$\mathrm{a}=\frac{1}{2},\mathrm{b}=\frac{-1}{(-1)\left(1\right)}=1,\mathrm{c}=\frac{-3}{(-2)(-1)}=-\frac{3}{2}\Rightarrow$$$$\mathrm{F}\left(\mathrm{x}\right)=\frac{1}{2\mathrm{x}}+\frac{1}{\mathrm{x}+1}-\frac{3}{2(\mathrm{x}+2)}\Rightarrow$$$$\sum _{\mathrm{k}=1}^{\mathrm{n}}\mathrm{F}\left(\mathrm{k}\right)=\frac{1}{2}\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{1}{\mathrm{k}}+\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{1}{\mathrm{k}+1}-\frac{3}{2}\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{1}{\mathrm{k}+2}$$$$\mathrm{we}\mathrm{have}\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{1}{\mathrm{k}}={\mathrm{H}}_{\mathrm{n}}$$$$\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{1}{\mathrm{k}+1}=\sum _{=2}^{\mathrm{n}+1}\frac{1}{\mathrm{k}}={\mathrm{H}}_{\mathrm{n}}-1+\frac{1}{\mathrm{n}+1}$$$$\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{1}{\mathrm{k}+2}=\sum _{\mathrm{k}=3}^{\mathrm{n}+2}\frac{1}{\mathrm{k}}={\mathrm{H}}_{\mathrm{n}}-\frac{3}{2}+\frac{1}{\mathrm{n}+1}+\frac{1}{\mathrm{n}+2}\Rightarrow$$$$\sum _{\mathrm{k}=1}^{\mathrm{n}}\mathrm{F}\left(\mathrm{k}\right)=\frac{1}{2}{\mathrm{H}}_{\mathrm{n}}+{\mathrm{H}}_{\mathrm{n}}-1+\frac{1}{\mathrm{n}+1}-\frac{3}{2}({\mathrm{H}}_{\mathrm{n}}-\frac{3}{2}+\frac{1}{\mathrm{n}+1}+\frac{1}{\mathrm{n}+2})$$$$=-1+\frac{1}{\mathrm{n}+1}+\frac{9}{4}-\frac{3}{2(\mathrm{n}+1)}-\frac{3}{2(\mathrm{n}+2)}\to \frac{9}{4}-1=\frac{5}{4}\Rightarrow$$$$\mathrm{S}=\frac{5}{4}$$

Commented by mathdanisur last updated on 30/Jun/21

$$coolthanksSir$$

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