Answer
Converges
Work Step by Step
We compare with: $\Sigma_{n=1}^\infty \dfrac{2^n+3^n}{3^n+4^n} \leq \Sigma_{n=1}^\infty \dfrac{(3^n+3^n)}{(3^n+4^n)} \leq \Sigma_{n=1}^\infty\dfrac{(3^n+3^n)}{4^{n}}$
This implies that $\Sigma_{n=1}^\infty\dfrac{(3^{n}+3^n)}{4^n}= (2)\Sigma_{n=1}^\infty (\dfrac{3}{4})^n$
Here, the series $\Sigma_{n=1}^\infty 2(\dfrac{3}{4})^n$ shows a geometric convergent series and has common ratio $r \lt 1$.
Hence, the series converges by the direct comparison test.