Chapter 10.4, Problem 33E

To determine

To find:The convergence or divergence of the series.

  ${\displaystyle \sum _{n=1}^{\infty }\frac{2^n}{3^n+1}}$

Answer to Problem 33E

The series ${\displaystyle \sum _{n=1}^{\infty }\frac{2^n}{3^n+1}}$ is convergent.

Explanation of Solution

Given information: ${\displaystyle \sum _{n=1}^{\infty }\frac{2^n}{3^n+1}}$

Calculation:We use the Ratio test.

  $\begin{array}{l}\underset{n\to \infty }{\lim }\frac{a_{n+1}}{a_n}\\ =\underset{n\to \infty }{\lim }\frac{\frac{2^{n+1}}{3^{n+1}+1}}{\frac{2^n}{3^n+1}}\\ =\underset{n\to \infty }{\lim }\frac{2^{n+1}}{3^{n+1}+1}\cdot \frac{3^n+1}{2^n}\\ =\underset{n\to \infty }{\lim }2\frac{3^n+1}{3^{n+1}+1}\\ =\underset{n\to \infty }{\lim }2\frac{1+\frac{1}{3^n}}{3+\frac{1}{3^n}}\\ =\frac{2}{3}\end{array}$

The series converges because the ratio $\frac{2}{3}<1$ .