Summation with two sequences

$\begin{aligned} \sum_{n=1}^\infty \frac{a_n}{b_n} & = \sum_{n=1}^\infty \frac{2^n(5n+18)}{3^nn(n+2)} \\ & = \sum_{n=1}^\infty \left[ \left(\frac{2}{3}\right)^n \left(\frac{5}{n+2} + 9\left(\frac{1}{n} - \frac{1}{n+2} \right) \right) \right] \\ & = \sum_{n=1}^\infty \left[ \left(\frac{2}{3}\right)^n \left(\frac{9}{n} - \frac{4}{n+2}\right) \right] \\ & = \sum_{n=1}^\infty \frac{2^n}{3^{n-2}n} - \sum_{n=1}^\infty \frac{2^{n+2}}{3^n(n+2)} \\ & = \sum_{n=1}^\infty \frac{2^n}{3^{n-2}n} - \sum_{n=3}^\infty \frac{2^{n}}{3^{n-2}n} \\ & = \sum_{n=1}^2 \frac{2^n}{3^{n-2}n} = \frac{2}{3^{-1}} + \frac{2^2}{3^{0}2} = 6 + 2 = \boxed{8} \end{aligned}$

$\begin{aligned} \dfrac{5n+18}{n(n+2)}&=\dfrac 9 n -\dfrac 4 {n+2}~~~~\text{Partial fraction for telescoping series.}\\ \therefore\sum_{n=1}^\infty \frac{a_n}{b_n} & = \sum_{n=1}^\infty \frac{2^n(5n+18)}{3^nn(n+2)} \\ & = \sum_{n=1}^\infty \left(\frac{2}{3}\right)^n \left(\frac{9}{n} - \frac{4}{n+2}\right) ...............(a)\\ & = \sum_{n=1}^\infty \frac{2^n}{3^{n-2}n} - \sum_{n=1}^\infty \frac{2^{n+2}}{3^n(n+2)} ..............(b) \\ & = \sum_{n=1}^2 \frac{2^n}{3^{n-2}n} +\sum_{n=3}^\infty \frac{2^n}{3^{n-2}n}- \sum_{n=1}^\infty \frac{2^{n+2}}{3^n(n+2)} \\ & = \sum_{n=1}^2 \frac{2^n}{3^{n-2}n} +\color{#3D99F6}{\sum_{n=1}^\infty \frac{2^{n+2} } {3^n(n+2)}- \sum_{n=1}^\infty \frac{2^{n+2}}{3^n(n+2)}} ..(c)\\ & = \sum_{n=1}^2 \frac{2^n}{3^{n-2}n} = \frac{2}{3^{-1}} + \frac{2^2}{3^{0}2} = 6 + 2 =~~~~~\color{#D61F06}{ \boxed{8}} \end{aligned} \\ \text{This is almost the same as the solution by Chew-Seong Cheong above.}\\\text{His conversion from } \displaystyle \sum_1~ to ~\sum_3 \text{ I feel is better than my } \sum_3 to \sum_1 \\ \text{in the last but one lines (c). Lines (a) and (b) I have taken from him}\\ to~ improve~ my~ solution. \\ \displaystyle \color{#EC7300} {\text{NOTE:- If the limit was finite in place of infinity say up to 2015,}\\ \text{we would have {(n+2)-n}= TWO more terms to add algebraically. }\\ - \sum_{n=2014}^{2015}\dfrac{2^{n+2}}{3^n(n+2)} }$