Answer
$1-\dfrac{1}{n}$. and sum $=1$
Work Step by Step
Consider $ \Sigma_{n=1}^{\infty} a_n=\Sigma_{n=1}^{\infty} (\dfrac{1}{n}-\dfrac{1}{n+1})$
and $s_n= 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-.....+\dfrac{1}{n-1}-\dfrac{1}{n}=1-\dfrac{1}{n}$
So, we get the nth partial sums is $(1-\dfrac{1}{n})$.
We can see that when $n \to \infty$ then $s_n \to 1$ .
Thus, the series converges.
Sum is equal to $(1-\lim\limits_{n \to \infty}\dfrac{1}{n})=1-0=1$.
