Answer
Converges
Work Step by Step
Consider $a_n=\dfrac{\sqrt n+1}{n+1}$
Here, $a_n$ is positive for all the values of $n$.
Now, $f(n)=\dfrac{\sqrt n+1}{n+1}$
and $f'(n)=-\dfrac{(\dfrac{1}{2 \sqrt n}) (n+1)-(1) (\sqrt n+1)}{(n+1)^2} =-\dfrac{1+\dfrac{\sqrt n }{2} -\dfrac{1}{2 \sqrt n}}{(n+1)^2} \leq 0$
The negative sign shows that, the sequence $a_n$ is not increasing.
we can see that $\lim\limits_{n \to \infty} a_n=\lim\limits_{n \to \infty}\dfrac{\sqrt n+1}{n+1}$
or, $\dfrac{1+1/\sqrt n}{\sqrt n+1/\sqrt n}=0$
Hence, by the Alternating Series Test, the series converges.