How can I use the telescoping technique to compute the following sum? I'm having issues getting started. I know the basic steps but I don't know how to perform them. I know I have to separate the fraction into A and B. After that I have to perform the sum but I'm not sure what comes next.
$$\sum_{i=3}^n \frac{1}{i(i+3)} $$
Thanks for any help!
On
$\displaystyle\sum_{i=3}^n \dfrac{1}{i(i+3)}=\dfrac{1}{3}\sum_{i=3}^n \left(\dfrac{1}{i}-\dfrac{1}{i+3}\right)=\dfrac{1}{3}\sum_{i=3}^n \left(\dfrac{1}{i}-\dfrac{1}{i+1}+\dfrac{1}{i+1}-\dfrac{1}{i+2}+\dfrac{1}{i+2}-\dfrac{1}{i+3}\right)$
$\displaystyle=\dfrac{1}{3}\sum_{i=3}^n \left(\dfrac{1}{i}-\dfrac{1}{i+1}\right)+\dfrac{1}{3}\sum_{i=3}^n \left(\dfrac{1}{i+1}-\dfrac{1}{i+2}\right)+\dfrac{1}{3}\sum_{i=3}^n \left(\dfrac{1}{i+2}-\dfrac{1}{i+3}\right)$
$=\dfrac{1}{3}\left(\dfrac{1}{3}-\dfrac{1}{n+1}+\dfrac{1}{4}-\dfrac{1}{n+2}+\dfrac{1}{5}-\dfrac{1}{n+3}\right)$
$=\dfrac{(n-2)(47n^2+196n+189)}{180(n+1)(n+2)(n+3)}$
$\sum_{i=3}^{n} \frac{1}{i(i+3)} = \frac{1}{3} \sum_{i=3}^{n} \frac{1}{i} - \frac{1}{i+3}$ via partial fractions.
To see what's going on, try writing out the first 4 or so terms and you'll quickly see what cancels out and what's left over...