Answer
Converges
Work Step by Step
Consider $u_n=\dfrac{1}{\sqrt n}$
Here, $u_n$ is positive for all the values of $n$.
Now, $f(n)=\dfrac{1}{\sqrt n}$
and $f'(n)=\dfrac{-1}{2\sqrt {n^3}} \lt 0$
The negative sign shows that, the sequence $u_n$ is not increasing.
Thus, $\lim\limits_{n \to \infty} u_n=\lim\limits_{n \to \infty}\dfrac{1}{\sqrt n}=0$
Hence, by the Alternating Series Test, the series converges.