6
$\begingroup$

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?

  • 9
    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.2011-10-25
  • 6
    +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.2011-10-25
  • 0
    Hints: You will find out whether the identity is useful or not if you write out the first few terms in full and look at what happens and whether terms do cancel. There is a little trick to do with powers of 2 which you might helpfully spot, which would make this more immediately obvious in your equality.2011-10-25
  • 0
    $\frac{167}{96}$2011-10-25
  • 1
    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.2011-10-25
  • 1
    Please do what Gerry is suggesting, if you have indeed a solution figured out.2011-10-25
  • 1
    @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.2011-10-26
  • 0
    @joriki, agreed.2011-10-26

1 Answers 1

5

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

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

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

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

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

  • 0
    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*}2011-10-26