Answer
The series is convergent and the sum of the series is $\frac{11}{6}$.
Work Step by Step
Define the $n-$th partial sum: $s_n=\sum_{i=1}^n\frac{3}{i(i+3)}$.
Simplify $s_n$:
$s_n=\sum_{i=1}^n\frac{3}{i(i+3)}$
$=\sum_{i=1}^n\frac{(i+3)-i}{i(i+3)}$
$=\sum_{i=1}^n(\frac{1}{i}-\frac{1}{i+3})$
$=\sum_{i=1}^n(\frac{1}{i}-\frac{1}{i+1}+\frac{1}{i+1}-\frac{1}{i+2}+\frac{1}{i+2}-\frac{1}{i+3})$
$=\sum_{i=1}^n(\frac{1}{i}-\frac{1}{i+1})+\sum_{i=1}^n(\frac{1}{i+1}-\frac{1}{i+2})+\sum_{i=1}^n(\frac{1}{i+2}-\frac{1}{i+3})$
$=\frac{1}{1}-\frac{1}{n+1}+\frac{1}{1+1}-\frac{1}{n+2}+\frac{1}{1+2}-\frac{1}{n+3}$
$=1+\frac{1}{2}+\frac{1}{3}-\frac{1}{n+1}-\frac{1}{n+2}-\frac{1}{n+3}$
$=\frac{11}{6}-\frac{1}{n+1}-\frac{1}{n+2}-\frac{1}{n+3}$
Find the sum of the series:
$\sum_{n=1}^\infty\frac{3}{n(n+3)}=\lim\limits_{n \to \infty }s_n=\lim\limits_{n \to \infty }(\frac{11}{6}-\frac{1}{n+1}-\frac{1}{n+2}-\frac{1}{n+3})=\frac{11}{6}-0-0-0=\frac{11}{6}$
So, the series is convergent and the sum of the series is $\frac{11}{6}$.
