Answer
Converges, by Th. 9.12
Work Step by Step
Th. 9.12 Direct Comparison Test:
Let $0 < a_{n} \leq b_{n}$ for all $n$.
1. If $\displaystyle \sum_{n=1}^{\infty}b_{n}$ converges, then $\displaystyle \sum_{n=1}^{\infty}a_{n}$ converges.
2. If $\displaystyle \sum_{n=1}^{\infty}a_{n}$ diverges, then $\displaystyle \sum_{n=1}^{\infty}b_{n}$ diverges.
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Observe that
$0 < \displaystyle \frac{1}{3n^{2}+2} < \frac{1}{3n^{2}} $ for $ n \geq 1$
$a_{n}=\displaystyle \frac{1}{3n^{2}+2},\qquad b_{n}=\frac{1}{3n^{2}}$
$ 0 < a_{n} \leq b_{n}$
(1). Since $\displaystyle \sum_{n=1}^{\infty}b_{n}=\frac{1}{3}\sum_{n=1}^{\infty}\frac{1}{n^{2}}$ converges
($p= 2 > 1,$ convergent p-series, Th.9.11),
then $\displaystyle \sum_{n=1}^{\infty}a_{n} =\displaystyle \sum_{n=1}^{\infty}\frac{1}{2n-1}$ converges