## University Calculus: Early Transcendentals (3rd Edition)

Since, we have $\Sigma_{n=1}^\infty \dfrac{2^n+3^n}{3^{n}+4^n} \leq \Sigma_{n=1}^\infty \dfrac{3^n+3^n}{3^{n}+4^n} \leq \Sigma_{n=1}^\infty\dfrac{3^{n}+3^n}{4^n}$ and $\Sigma_{n=1}^\infty\dfrac{3^{n}+3^n}{4^n}= \Sigma_{n=1}^\infty 2(\dfrac{3}{4})^n$ The series $\Sigma_{n=1}^\infty 2(\dfrac{3}{4})^n$ is a convergent geometric series with common ratio $r \lt 1$. Thus, the given series is convergent as well.