Evaluate

$\displaystyle \sum_{k=2}^{\infty}\,\,(\frac{2}{3^{k}} + \frac{1}{2^{k}})$

So I separated the two sums and also change the dummy variable using $\displaystyle n=k-2$ so I can eventually have two geometric series:

$\displaystyle 2\sum_{n=0}^{\infty}\,\,(\frac{1}{3})^{n+2} + \sum_{n=0}^{\infty}\,\,(\frac{1}{2})^{n+2}$

Pulling out the constants within the sums, I can use $\displaystyle \frac{a}{1-r}$ for both sums:

$\displaystyle \frac{\frac{2}{9}}{1-\frac{1}{3}} + \frac{\frac{1}{4}}{1-\frac{1}{2}}$

Simplifying, I get:

$\displaystyle \frac{2}{9}\cdot\frac{3}{2}+\frac{1}{4}\cdot\frac{ 2}{1}=\frac{1}{3}+\frac{1}{2}=\frac{5}{6}$

However, my book states that the answer is $\displaystyle \frac{6}{5}$. Did I do something wrong, or is the book's answer key wrong? Any help would be appreciated, thanks.