Answer
Converges by Alternating Series (Theorem 9.14)
Work Step by Step
Rewrite the summation, $\Sigma^{\infty}_{n=1} (-1)^{n+1} \frac{\sqrt n}{n+2}$
$\lim\limits_{n \to \infty} \frac{\sqrt n}{n+2} = \frac{\infty}{\infty} $
Use L'Hopital's Rule
$\lim\limits_{n \to \infty} \frac{1}{2 \sqrt n} =0$
Satisfies the first test, Next see if $a_{n+1} \leq a_n$
By checking a few numbers one can see that
$a_{n+1} = \frac{\sqrt {n+1}}{n+3} \leq \frac{\sqrt n}{n+2} = a_n$
Both conditions are satisfied so the Series Converges by Alternating Series (Theorem 9.14)
