Answer
Divergent
Work Step by Step
$1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{9}+.....=\Sigma_{n=1}^{\infty} \frac{1}{2n-1}$
$\int_{1}^{\infty}\frac{1}{2x-1}=\lim\limits_{t \to \infty}\int_{1}^{t} \frac{1}{2x-1}=\lim\limits_{t \to \infty}[\frac{1}{2}ln(2x-1)]^{t}_{1}=\infty$
Hence, the given series is divergent.