Division in Summations Suppose $a_n=\dfrac{2^n}{n(n+2)}$ and $b_n=\dfrac{3^n}{5n+18}$. 
I need to find the value of:
$$\displaystyle\sum_{n=1}^{\infty}\dfrac{a_n}{b_n}$$
I think this problem is meant for me to compute each sum differently and then divide. Is this some property of summations that we need to utilize here?

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