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Question Number 194445 by tri26112004 last updated on 06/Jul/23

$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n^2(n+a)}=¿$$$$(a\ne 0)$$

Answered by mr W last updated on 08/Jul/23

$$\frac{An+B}{n^2}+\frac{C}{n+a}=\frac{(A+C)n^2+(aA+B)n+aB}{n^2(n+a)}$$$$aB=1\Rightarrow B=\frac{1}{a}$$$$aA+B=0\Rightarrow A=-\frac{B}{a}=-\frac{1}{a^2}$$$$A+C=0\Rightarrow C=-1=\frac{1}{a^2}$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n^2(n+a)}$$$$=\underset{n=1}{\sum ^{\mathrm{\infty }}}(\frac{B}{n^2}+\frac{A}{n}+\frac{C}{n+a})$$$$=B\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n^2}+\underset{n=1}{\sum ^{\mathrm{\infty }}}\left(\frac{A}{n}\right)+\underset{n=1}{\sum ^{\mathrm{\infty }}}\left(\frac{C}{n+a}\right)$$$$=B\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n^2}+\underset{n=1}{\sum ^a}\left(\frac{A}{n}\right)+\underset{n=a+1}{\sum ^{\mathrm{\infty }}}\left(\frac{A}{n}\right)+\underset{n=1}{\sum ^{\mathrm{\infty }}}\left(\frac{C}{n+a}\right)$$$$=B\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n^2}+A\underset{n=1}{\sum ^a}\frac{1}{n}+\underset{n=1}{\sum ^{\mathrm{\infty }}}\left(\frac{A}{n+a}\right)+\underset{n=1}{\sum ^{\mathrm{\infty }}}\left(\frac{C}{n+a}\right)$$$$=B\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n^2}+A\underset{n=1}{\sum ^a}\frac{1}{n}+\underset{n=1}{\sum ^{\mathrm{\infty }}}\left(\frac{A+C}{n+a}\right)$$$$=B\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n^2}+A\underset{n=1}{\sum ^a}\frac{1}{n}$$$$=\frac{1}{a}\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n^2}-\frac{1}{a^2}\underset{n=1}{\sum ^a}\frac{1}{n}$$$$=\frac{{\pi }^2}{6a}-\frac{1}{a^2}\underset{n=1}{\sum ^a}\frac{1}{n}$$

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