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