n-1-1-n-2-n-a-a-0-

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

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

(a \neq 0)

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

\frac{A n + B}{n^{2}} + \frac{C}{n + a} = \frac{(A + C) n^{2} + (a A + B) n + a B}{n^{2} (n + a)}

a B = 1 \Rightarrow B = \frac{1}{a}

a A + 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^{\infty}} \frac{1}{n^{2} (n + a)}

= \underset{n = 1}{\sum^{\infty}} (\frac{B}{n^{2}} + \frac{A}{n} + \frac{C}{n + a})

= B \underset{n = 1}{\sum^{\infty}} \frac{1}{n^{2}} + \underset{n = 1}{\sum^{\infty}} (\frac{A}{n}) + \underset{n = 1}{\sum^{\infty}} (\frac{C}{n + a})

= B \underset{n = 1}{\sum^{\infty}} \frac{1}{n^{2}} + \underset{n = 1}{\sum^{a}} (\frac{A}{n}) + \underset{n = a + 1}{\sum^{\infty}} (\frac{A}{n}) + \underset{n = 1}{\sum^{\infty}} (\frac{C}{n + a})

= B \underset{n = 1}{\sum^{\infty}} \frac{1}{n^{2}} + A \underset{n = 1}{\sum^{a}} \frac{1}{n} + \underset{n = 1}{\sum^{\infty}} (\frac{A}{n + a}) + \underset{n = 1}{\sum^{\infty}} (\frac{C}{n + a})

= B \underset{n = 1}{\sum^{\infty}} \frac{1}{n^{2}} + A \underset{n = 1}{\sum^{a}} \frac{1}{n} + \underset{n = 1}{\sum^{\infty}} (\frac{A + C}{n + a})

= B \underset{n = 1}{\sum^{\infty}} \frac{1}{n^{2}} + A \underset{n = 1}{\sum^{a}} \frac{1}{n}

= \frac{1}{a} \underset{n = 1}{\sum^{\infty}} \frac{1}{n^{2}} - \frac{1}{a^{2}} \underset{n = 1}{\sum^{a}} \frac{1}{n}

= \frac{π^{2}}{6 a} - \frac{1}{a^{2}} \underset{n = 1}{\sum^{a}} \frac{1}{n}

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