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