Answer
Convergent
Work Step by Step
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.