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?
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?
$\begin{align*} \frac{4n-3}{n(n^2-4)}=\frac{3}{4n}-\frac{11}{8(n+2)}+\frac{5}{8(n-2)} \end{align*}$
$\begin{align*} \frac{4n-3}{n(n^2-4)}=\frac{1}{8}(6(\frac{1}{n}-\frac{1}{n+2})+5(\frac{1}{n-2}-\frac{1}{n+2})) \end{align*}$
$\begin{align*} \sum_{n=3}^{m}\frac{1}{n}-\frac{1}{n+2}=\frac{7}{12}-\frac{1}{m+1}-\frac{1}{m+2}\rightarrow7/12\end{align*}$
$\begin{align*} \sum_{n=3}^{m}\frac{1}{n-2}-\frac{1}{n+2}=\frac{25}{12}-\frac{1}{m-1}-\frac{1}{m}-\frac{1}{m+1}-\frac{1}{m+2}\rightarrow25/12 \end{align*}$
$\begin{align*} \sum_{n=3}^{\infty} \frac{4n-3}{n(n^2-4)}=\frac{1}{8}(6\times7/12+5\times25/12)=\frac{167}{96} \end{align*}$