Answer
Converges
Work Step by Step
Let us consider $u_n=\dfrac{\sqrt n+1}{n+1}$
Now, $f(n)=\dfrac{\sqrt n+1}{(n+1)}$
and $f'(n)=-\dfrac{(\dfrac{1}{2 \sqrt n}) (n+1)-(\sqrt n+1)(1)}{(n+1)^2} =-\dfrac{1+(\sqrt n/2) -(1/2\sqrt n)}{(n+1)^2} \leq 0$
Here,the negative sign shows that, the sequence $a_n$ is decreasing.
Now, $\lim\limits_{n \to \infty} u_n=\lim\limits_{n \to \infty}\dfrac{\sqrt n+1}{(n+1)} \implies \dfrac{1+\dfrac{1}{\sqrt n}}{\sqrt n+\dfrac{1}{\sqrt n}}=0$
Hence, the series converges by the Alternating Series Test.
