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

Consider $u_n=\dfrac{(2n+3)(2^n+3)}{3^n+2}$ Here,$\Sigma_{n=1}^\infty \dfrac{(2n+3)(2^n+3)}{3^n+2} \leq \Sigma_{n=1}^\infty\dfrac{(2n+3)(2^n+3)}{3^n}$ Suppose $\Sigma_{n=1}^\infty a_n=\Sigma_{n=1}^\infty \dfrac{(2n+3)(2^n+3)}{3^n}$ Now, $l=\lim\limits_{n \to \infty} |\dfrac{a_{n+1}}{a_{n}} |=\lim\limits_{n \to \infty}|\dfrac{\dfrac{(2(n+1)+3))(2^{n+1}+3)}{3^{n+1}}}{\dfrac{(2n+3)(2^n+3)}{3^n}}|$ Thus, we have $l=(\dfrac{1}{3})\lim\limits_{n \to \infty}|(\dfrac{2n+5}{2n+3})|\lim\limits_{n \to \infty}|\dfrac{(2)(2^n+3)}{2^n+3})|=(\dfrac{1}{3})\lim\limits_{n \to \infty}|\dfrac{2+5/n}{2+3/n}|(2)$ So, $l=\dfrac{2}{3} \lt 1$ Thus, the series $\Sigma_{n=1}^\infty a_n=\Sigma_{n=1}^\infty \dfrac{(2n+3)(2^n+3)}{3^n}$ converges. Hence, the series Converges by the direct comparison test.