Proof that $\sum_{n=1}^{\infty}2^{-3n} = \frac{1}{7}$ Exactly as it says in the title. A proof that:
$$\sum_{n=1}^{\infty}2^{-3n} = \frac{1}{7}$$
I first realised this when I attempted to find the binary representation of $\frac{1}{7}$ and found that it was $1.\overline{001}$  (where the line denotes the recurring part). Since every bit at a position which is a multiple of $-3$ is a $1$, I derived that series. I would like a more mathematical proof; preferably as easy to understand as possible.
EDIT: The series given actually equals $\frac{8}{7}$, I made an error.

Here is easier solution...
$A=\frac {1}{8} + \frac {1}{64} + ..... = \frac {1}{8} + \frac {1}{8} \cdot  (\frac {1}{8} + \frac {1}{64} + .....) = \frac {1}{8} + \frac {A}{8} $
$ \frac {7A}{8} = \frac {1}{8} $
$ A = \frac {1}{7} $
