Answer
Converges
Work Step by Step
Given: $\Sigma _{n=1}^{\infty}\frac{9^{n}}{3+10^{n}}$
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 $|r|\lt 1$.
$\Sigma _{n=1}^{\infty}\frac{9^{n}}{3+10^{n}}\lt\Sigma _{n=1}^{\infty}\frac{9^{n}}{10^{n}}$
$=\Sigma _{n=1}^{\infty}(\frac{9}{10})^{n}$
$=\Sigma _{n=1}^{\infty}(\frac{9}{10})(\frac{9}{10})^{n-1}$
A geometric series with common ratio $r=\frac{9}{10}$ is converging.
We know that a series less than a converging series also converges.
