Multivariable Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 0-53849-787-4
ISBN 13: 978-0-53849-787-9

Chapter 11 - Infinite Sequences and Series - 11.1 Exercises: 28

Answer

The series converges.

Work Step by Step

$a_n=\frac{3^{n+2}}{5^n}$ 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.
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