# Chapter 9 - Section 9.2 - Infinite Series - Exercises - Page 498: 64

Diverges

#### Work Step by Step

The sum of a geometric series can be found as: $S=\dfrac{a}{1-r}$ Since, we have $\sum_{n =1}^{ \infty}\dfrac{2^n+4^n}{3^n+4^n}=\sum_{n =1}^{ \infty}\dfrac{(\dfrac{1}{2})^n+1}{(\dfrac{3}{4})^n+1}=\dfrac{1}{1}= 1\ne 0$ Thus, the given series $\sum_{n =1}^{ \infty}\dfrac{2^n+4^n}{3^n+4^n}$ diverges as per the nth term integral test.

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