$\displaystyle \sum_{k = 0}^\infty \frac{3^{k-1}}{4^{3k+1}}$

$\displaystyle \sum_{k = 0}^\infty \frac{3^{k}*3^{-1}}{4^{3k}*4^{1}}$

$\displaystyle \sum_{k = 0}^\infty \frac{3^{k}}{(4^{3})^{k}*4^{1}*3^{1}}$

$\displaystyle (\frac{1}{12})\sum_{k = 0}^\infty \frac{3^{k}}{64^{k}}$

$\displaystyle (\frac{1}{12})\sum_{k = 0}^\infty (\frac{3}{64})^{k}$

So it's a geometric series with $\displaystyle a = 1$ and $\displaystyle r = 3/64$. So $\displaystyle \frac{1}{12}*\frac{1}{1 - \frac{3}{64}}$ = $\displaystyle \frac{16}{183}$ ?

The solutions manual had $\displaystyle \frac{3}{15,616}$. Am I doing this right?