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I would like to find the exact value of the series

$$\begin{align*} \sum_{n=3}^{\infty} \frac{4n-3}{n(n^2-4)} \end{align*}$$

which is certainly a telescoping series. Do you have any idea of telescopic cancelling?

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It's somewhat surprising that you know what a telescoping series is and are able to identify one and to find this expression for the summand, but then don't see how to use it. Perhaps you should tell us more about what you know and what it is you're finding difficult. –  joriki Oct 25 '11 at 18:37
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+1, you are completely on the right track. Perhaps rewriting the right hand side of the above equality will make things clearer: $$\frac{1}{8}\left( \frac{6}{n} -\frac{11}{n+2} +\frac{5}{n-2}\right).$$ Notice that the numerators add to $0$, which is perfect, and tells us a little more about how the telescoping will occur. –  Eric Naslund Oct 25 '11 at 18:42
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Instead of editing the answer into the question, you can post it as an answer and then accept it. I gather this will make the software happy. –  Gerry Myerson Oct 25 '11 at 21:55
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Please do what Gerry is suggesting, if you have indeed a solution figured out. –  J. M. Oct 25 '11 at 22:48
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@Gerry: It doesn't just make the software happy; it makes everyone happy who doesn't encounter this question as an unanswered question and spends time reading it before noticing that it's been answered. –  joriki Oct 26 '11 at 2:35

1 Answer 1

up vote 3 down vote accepted

$$\begin{align*} \frac{4n-3}{n(n^2-4)}=\frac{3}{4n}-\frac{11}{8(n+2)}+\frac{5}{8(n-2)} \end{align*}$$

$$\begin{align*} \frac{4n-3}{n(n^2-4)}=\frac{1}{8}(6(\frac{1}{n}-\frac{1}{n+2})+5(\frac{1}{n-2}-\frac{1}{n+2})) \end{align*}$$

$$\begin{align*} \sum_{n=3}^{m}\frac{1}{n}-\frac{1}{n+2}=\frac{7}{12}-\frac{1}{m+1}-\frac{1}{m+2}\rightarrow7/12\end{align*}$$

$$\begin{align*} \sum_{n=3}^{m}\frac{1}{n-2}-\frac{1}{n+2}=\frac{25}{12}-\frac{1}{m-1}-\frac{1}{m}-\frac{1}{m+1}-\frac{1}{m+2}\rightarrow25/12 \end{align*}$$

$$\begin{align*} \sum_{n=3}^{\infty} \frac{4n-3}{n(n^2-4)}=\frac{1}{8}(6\times7/12+5\times25/12)=\frac{167}{96} \end{align*}$$

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Please, How was the sum in the LHS of 3rd step calculated to be equal to the RHS-What is m? \begin{align*} \sum_{n=3}^{m}\frac{1}{n}-\frac{1}{n+2}=\frac{7}{12}-\frac{1}{m+1}-\frac{1}{m+2}‌​\rightarrow7/12\end{align*} –  Emmad Kareem Oct 26 '11 at 8:00

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