Answer
diverges by comparison with a divergent series
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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$b_{n}=\displaystyle \frac{1}{2n-1}$
Observe that
$\displaystyle \frac{1}{2n-1} > \frac{1}{2n} > 0$ for $ n \geq 1$
$a_{n}=\displaystyle \frac{1}{2n},\qquad 0 < a_{n} \leq b_{n}$
(2). Since $\displaystyle \sum_{n=1}^{\infty}a_{n}=\frac{1}{2}\sum_{n=1}^{\infty}\frac{1}{n}$ diverges (divergent p-series, Th.9.11)),
then $\displaystyle \sum_{n=1}^{\infty}b_{n} =\displaystyle \sum_{n=1}^{\infty}\frac{1}{2n-1}$ diverges.
