Answer
The series is divergent.
Work Step by Step
$a_{n}=\frac{n}{\sqrt (n^{2}+4)}$
partial sum:
$s_{n} = a_{1} + a_{2}+...+a_{n}$
$a_{n}$ (blue dots)
$s_{n}$ (red dots}
$n=1$ $s_{1}=0.4472$ $a_{1}=0.4472$
$n=2$ $s_{2}=1.1543$ $a_{2}=0.7071$
$n=3$ $s_{3}=1.9864$ $a_{3}=0.8321$
$n=4$ $s_{4}=2.8808$ $a_{4}=0.8944$
$n=5$ $s_{5}=3.0893$ $a_{5}=0.9285$
$n=6$ $s_{6}=4.7580$ $a_{6}=0.9487$
$n=7$ $s_{7}=5.7195$ $a_{7}=0.9615$
$n=8$ $s_{8}=6.6896$ $a_{8}=0.9701$
$n=9$ $s_{9}=7.6658$ $a_{9}=0.9762$
$n=10$ $s_{10}=8.6464$ $a_{10}=0.9806$
The sequence terms $a_{n}$ appear to approach 1
Check limit $L$
$L = \lim\limits_{n \to \infty} \frac{n}{\sqrt (n^{2}+4)}$
$L^{2} = \lim\limits_{n \to \infty} \frac{n^{2}}{n^{2}+4} \times \frac{\frac{1}{n^{2}}}{\frac{1}{n^{2}}}$
$L^{2}=\lim\limits_{n \to \infty} \frac{1}{1+\frac{4}{n^{2}}} = \frac{1}{1+0} = 1$
$L=1 \ne 0$
The limit is not zero, so the series is divergent
