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Question Number 108415 by mathdave last updated on 16/Aug/20

Commented by bobhans last updated on 17/Aug/20

$$\frac{1}{n(n+2)(n+4)}=\frac{p}{n}+\frac{q}{n+2}+\frac{r}{n+4}$$$$1=p(n+2)(n+4)+qn(n+4)+rn(n+2)$$$$n=0\Rightarrow 1=8p\to p=\frac{1}{8}$$$$n=-2\Rightarrow 1=-4q\to q=-\frac{1}{4}$$$$n=-4\Rightarrow 1=8r\to r=\frac{1}{8}$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{8n}-\frac{1}{4(n+2)}+\frac{1}{8(n+4)}=I+J+H$$$$H=\frac{1}{8}\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n+4}=\frac{1}{8}(\frac{1}{5}+\frac{1}{6}+\underset{n=3}{\sum ^{\mathrm{\infty }}}\frac{1}{n+4})$$$$(\ast )\frac{1}{8}\underset{n=3}{\sum ^{\mathrm{\infty }}}\frac{1}{n+4}=\frac{1}{8}\underset{n=6}{\sum ^{\mathrm{\infty }}}\frac{1}{n+1}$$$$(\ast \ast )\frac{1}{8}\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n}=\frac{1}{8}(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\underset{n=6}{\sum ^{\mathrm{\infty }}}\frac{1}{n})$$$$$$

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