## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 11 - Infinite Series - Chapter Review Exercises - Page 592: 51

converges

#### Work Step by Step

Given $$\sum_{n=1}^{\infty} \frac{2^{n}+n}{3^{n}-2},\ \ \ \ b_{n}=\left(\frac{2}{3}\right)^{n}$$ We use the limit comparison test with $b_n$ given above: \begin{align*} \lim _{n \rightarrow \infty} \frac{a_{n}}{b_{n}}&=\lim _{n \rightarrow \infty} \frac{2^{n}+n}{3^{n}-2} \cdot \frac{3^{n}}{2^{n}}\\ &=\lim _{n \rightarrow \infty} \frac{6^{n}+n 3^{n}}{6^{n}-2^{n+1}}\\ &=\lim _{n \rightarrow \infty} \frac{1+n\left(\frac{1}{2}\right)^{n}}{1-2\left(\frac{1}{3}\right)^{n}}\\ &=1 \end{align*} Since $\sum \left(\frac{2}{3}\right)^{n}$ is a geometric convergent series, then $\sum_{n=1}^{\infty} \frac{2^{n}+n}{3^{n}-2}$ also converges.

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