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: $\Sigma_{n=1}^{\infty}(-1)^{n+1}\frac{ n^{2}}{n^{3}+4}$
In the given problem, $b_{n}=\frac{ n^{2}}{n^{3}+4}$
which satisfies both conditions of Alternating Series Test as follows:
1. $b_{n}=\frac{ n^{2}}{n^{3}+4}$ is decreasing because the denominator is increasing.
2. $\lim\limits_{n \to \infty}b_{n}=\lim\limits_{n \to \infty}\frac{ n^{2}}{n^{3}+4}$
$=\lim\limits_{n \to \infty}\frac{\frac{ n^{2}}{n^{3}}}{\frac{n^{3}}{n^{3}}+\frac{4}{n^{3}}}$
$=\lim\limits_{n \to \infty}\frac{\frac{ 1}{n}}{1+\frac{4}{n^{3}}}$
$=\frac{0}{1+0}$
$=0$
Hence, the given series is convergent by Alternating Series Test.