Answer
-1
Work Step by Step
Consider $s_n=\dfrac{-2}{n(n+1)}$
Re-write the given series as: $s_n=\dfrac{-2}{n(n+1)}=(\dfrac{-2}{2}+\dfrac{2}{3})+(\dfrac{-2}{3}+\dfrac{2}{4})+(\dfrac{-2}{4}+\dfrac{2}{5})+(\dfrac{-2}{5}+\dfrac{2}{6})...+(\dfrac{-2}{n}+\dfrac{2}{n+1})$
or,
$s_n=\dfrac{-2}{2}+\dfrac{2}{n+1}$
Apply limits, we get:
$\lim\limits_{n \to \infty}s_n=\lim\limits_{n \to \infty}\dfrac{-2}{2}+\dfrac{2}{n+1}$
After simplifications, we get
$\lim\limits_{n \to \infty}s_n=\lim\limits_{n \to \infty}(-1+\dfrac{2}{n+1})$
$\lim\limits_{n \to \infty}s_n=\lim\limits_{n \to \infty}(-1+\dfrac{2}{\infty})$
$\lim\limits_{n \to \infty}s_n=-1$
