Answer
Divergent
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: $\Sigma_{n=1}^{\infty}(-1)^{n}\frac{3n-1}{2n+1}$
In the given problem, $b_{n}=\frac{3n-1}{2n+1}$
1. Test the first condition of the AST as: $b_{n+1}\leq b_n$
$\frac{3(n+1)-1}{2(n+1)+1}\leq \frac{3n-1}{2n+1}$
$\frac{3n+2}{2n+3}\leq \frac{3n-1}{2n+1}$
$(3n+2)(2n+1)\leq (3n-1)(2n+3)$
$2\leq -3$
which is not true, so AST first condition fails.
2. $\lim\limits_{n \to \infty}b_{n}=\lim\limits_{n \to \infty}\frac{3n-1}{2n+1}=\lim\limits_{n \to \infty}\frac{3-1/n}{2+1/n}=\frac{3}{2}\ne 0$
which means that the series diverges by the Test of Divergence.
