Answer
Converges
Work Step by Step
As we are given that $\dfrac{(2n+3)(2^n+3)}{3^n+2}$
Apply direct comparison test.
we can see that $\Sigma_{n=1}^\infty \dfrac{(2n+3)(2^n+3)}{3^n+2} \leq \Sigma_{n=1}^\infty\dfrac{(2n+3)(2^n+3)}{(3^n)}$
Let us consider $\Sigma_{n=1}^\infty a_n=\Sigma_{n=1}^\infty \dfrac{(2n+3)(2^n+3)}{(3^n)}$
$\lim\limits_{n \to \infty} |\dfrac{a_{n+1}}{a_{n}} |=\lim\limits_{n \to \infty}|\dfrac{\dfrac{(2(n+1)+3))(2^{n+1}+3)}{3^{n+1}}}{\dfrac{(2n+3)(2^n+3)}{3^n}}|$
$(\dfrac{1}{3})\lim\limits_{n \to \infty}|(\dfrac{2n+5}{2n+3})|\lim\limits_{n \to \infty}|\dfrac{(2)(2^n+3)}{2^n+3})|=\dfrac{1}{3} \cdot \lim\limits_{n \to \infty}|\dfrac{2+\dfrac{5}{n}}{2+\dfrac{3}{n}}| \cdot 2=\dfrac{2}{3} \lt 1$
This shows that the series $\Sigma_{n=1}^\infty a_n=\Sigma_{n=1}^\infty \dfrac{(2n+3)(2^n+3)}{3^n}$ converges.
Thus, the series converges by the direct comparison test.