Answer
Convergent
Work Step by Step
Since the terms of the series involve $(-1)^n$, we need to use the Alternating Series Test.
For $n\geq 1$,
$2\leq n^2+n$ (Add with $n^3+n^2+2n$)
$n^3+n^2+2n+2\leq n^3+2n^2+3n$ (Factorize)
$(n+1)(n^2+2)\leq n(n^2+2n+3) $
$(n+1)(n^2+2)\leq n((n+1)^2+2)$
$\frac{n+1}{(n+1)^2+2}\leq \frac{n}{n^2+2}$
From the last equation, we have $b_{n+1}\leq b_n$ for all $n$.
In addition, we know $\lim b_n=0$.
Thus, it follows by the Alternating Series Test that $\sum_{n=1}^\infty (-1)^n\frac{n}{n^2+2}$ is convergent.
