## Multivariable Calculus, 7th Edition

$a_n=\frac{3^{n+2}}{5^n}$ Use the properties of exponents to split $3^{n+2}=3^n3^2$. Since $3^2=9$ we can simplify the expression. $\frac{3^23^n}{5^n}=9(\frac{3}{5})^n$, so $\lim\limits_{n \to \infty}a_n=9\lim\limits_{n\to\infty}(\frac{3}{5})^n\to$ $9\times0=0$ since this is a Geometric Series with $[|r|=(\frac{3}{5})<1]$. The series is convergent.