sum of the series $\frac{2^n+3^n}{6^n}$ from $n=1$ to $\infty$ Find the sum of the series $\sum_{n=1}^{\infty} \frac{2^n+3^n}{6^n}=?$
My thoughts: find $\sum_{n=1}^{\infty} 2^n$, $\sum_{n=1}^{\infty} 3^n$ and $\sum_{n=1}^{\infty} 6^n$ (although I don't know how yet...)
Then, $\sum_{n=1}^{\infty} \frac{2^n+3^n}{6^n}= \frac{\sum_{n=1}^{\infty} 2^n}{\sum_{n=1}^{\infty} 3^n} + \frac{\sum_{n=1}^{\infty} 3^n}{\sum_{n=1}^{\infty} 6^n}$, am I on the right track? I really don't know how else to solve it. 
Thanks.

$$\sum_{i=1}^\infty \frac{2^n + 3^n}{6^n}=\sum_{i=1}^\infty (\frac{2}{6})^n + \sum_{i=1}^\infty (\frac{3}{6})^n=\frac{1}{2}+1=\frac{3}{2}$$
sums from forumla of geometric series $1/3^n$ and  $1/2^n$
