## Calculus: Early Transcendentals 8th Edition

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.