Answer
Diverges
Work Step by Step
Given: $\Sigma _{n=1}^{\infty}\frac{6^{n}}{5^{n}-1}$
Any series of the form $\Sigma _{n=1}^{\infty}ar^{n-1}$ is called a geometric series.
A geometric series with common ratio $r$ converges only when $|r|\lt 1$.
$\Sigma _{n=1}^{\infty}\frac{6^{n}}{5^{n}-1}\gt\Sigma _{n=1}^{\infty}\frac{6^{n}}{5^{n}}$
$=\Sigma _{n=1}^{\infty}(\frac{6}{5})^{n}$
$=\Sigma _{n=1}^{\infty}(\frac{6}{5})(\frac{6}{5})^{n-1}$
A geometric series with common ratio $r=\frac{6}{5}$ is converging.
We know that a series greater than a diverging series also diverges.