Answer
Diverges
Work Step by Step
Let $u_n=\dfrac{\sqrt n+1}{\sqrt {n^2+3}}$
When the numerator increases, the value of fraction is always increasing and also, when the denominator decreases, the value of fraction is always increasing.
$a_n \geq \dfrac{\sqrt n}{\sqrt {n^2+3}}$
$\implies \dfrac{\sqrt n}{\sqrt {(n^2+3)}} \geq \dfrac{\sqrt n}{\sqrt{n^2+3n}}=\dfrac{1}{\sqrt {(n+3)}}$
Thus, $u_n \geq \dfrac{1}{\sqrt {(n+3)}}$
So, the series $\Sigma_{n=1}^\infty \dfrac{1}{\sqrt {n}}$ diverges.
Hence, the series diverges by the comparison test.