Answer
Convergent
Work Step by Step
Alternating series test:
Suppose that we have series $\Sigma a_n$, such that $a_{n}=(-1)^{n}b_n$ or $a_{n}=(-1)^{n+1}b_n$, where $b_n\geq 0$ for all $n$.
Then if the following two condition are satisfied the series is convergent.
1. $\lim\limits_{n \to \infty}b_{n}=0$
2. $b_{n}$ is a decreasing sequence.
Given: $\Sigma_{n=1}^{\infty}(-1)^{n}\frac{\sqrt n}{2n+3}$
In the given problem, $b_{n}=\frac{\sqrt n}{2n+3}$
which satisfies both conditions of Alternating Series Test as follows:
1. Let $f(x)=\frac{\sqrt n}{2n+3}$ and $f'(x)=\frac{3-2x}{2\sqrt x (2x+3)^{2}} \lt 0$
Thus, $f(x)$ is deceasing , so $b_{n}=\frac{\sqrt n}{2n+3}$ is decreasing because the denominator is increasing.
2. $\lim\limits_{n \to \infty}b_{n}=\lim\limits_{n \to \infty}\frac{\sqrt n}{2n+3}$
$=\lim\limits_{n \to \infty}\frac{\frac{\sqrt n}{n}}{2+\frac{3}{n}}$
$=\lim\limits_{n \to \infty}\frac{\frac{1}{\sqrt n}}{2+\frac{3}{n}}$
$=0$
Hence, the given series is convergent by Alternating Series Test.