Answer
Divergent
Work Step by Step
Consider $f(x)=\dfrac{1}{2x-1}$
This is positive, continuous and decreasing for $x \geq 0$
Now, take the integral test to find the convergence and divergence for the sequence.
We have $\lim\limits_{a \to \infty} \int_1^a \dfrac{1}{ 2x-1}dx= \lim\limits_{a \to \infty}\dfrac{1}{2} [\ln (2a-1)]=\infty$
Hence, the sequence $\Sigma_{n=0}^\infty \dfrac{1}{ 2n-1}$ is Divergent.
