Chapter 10.4, Problem 57E

To determine

To find: the correct option for this series.

Answer to Problem 57E

  $\frac{2}{3}$

Explanation of Solution

Given:

  $\underset{n\to \infty }{\lim }\frac{a_{n+1}}{a_n}$ For the series ${\displaystyle \sum _{n=0}^{\infty }\frac{2^n}{{\left(-3\right)}^n}}$

Calculation:

  $\underset{n\to \infty }{\lim }\frac{\frac{2^{n+1}}{{\left(-3\right)}^{n+1}}}{\frac{2^n}{{\left(-3\right)}^n}}$

The problem given for $\underset{n\to \infty }{\lim }\frac{a_{n+1}}{a_n}$

  $a_{n+1}=\frac{2^{n+1}}{{\left(-3\right)}^{n+1}}$

  $a_n=\frac{2^n}{{\left(-3\right)}^n}$

  $\underset{n\to \infty }{\lim }=\frac{2^{n+1}}{{\left(-3\right)}^{n+1}}\cdot \frac{{\left(-3\right)}^n}{2^n}$

  $\underset{n\to \infty }{\lim }=\frac{2}{-3}=-\frac{2}{3}$

Hence, the correct option is (b)