Answer
Divergent
Work Step by Step
Let us consider $f(x)=\dfrac{1}{2x-1}$
Here, the function $f(x)$ is positive, continuous and decreasing for $x \geq 0$
Then, $\lim\limits_{k \to \infty} \int_1^k \dfrac{1}{ 2x-1}dx= \lim\limits_{k \to \infty}(\dfrac{1}{2}) [\ln (2k-1)-\ln 1]=\infty$
Hence, the sequence $\Sigma_{n=0}^\infty \dfrac{1}{ 2n-1}$ is Divergent.
