Answer
Diverges
Work Step by Step
Consider $a_n=\dfrac{\sqrt n+1}{\sqrt {n^2+3}}$
When we decrease the numerator, the value of the fraction will always decrease:
$a_n \geq \dfrac{\sqrt n}{\sqrt {n^2+3}}$
and when we increase the denominator, the value of the fraction will always increase:
$ \dfrac{\sqrt n}{\sqrt {n^2+3}} \geq \dfrac{\sqrt n}{\sqrt{n^2+3n}}=\dfrac{1}{\sqrt {n+3}}$
we can see that $a_n \geq \dfrac{1}{\sqrt {n+3}}$
and we know that $\Sigma_{n=1}^\infty \dfrac{1}{\sqrt {n}}$ diverges.
Thus the given series diverges by the comparison test.
