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.
 A: This can be rewritten into a geometrices series, because
$$2^{-3n} = (2^{-3})^{n} = \left(\frac{1}{8}\right)^n. $$
Then we can use the geometric series formula for $|x|<1$ that says
$$\sum\limits_{n=0}^\infty x^n  = \frac{1}{1-x}$$
where $x = 1/8$.
A: Use the expression for a geometric series of common ratio $\vert r \vert \lt 1$ and first term $2^{-3}$:
$$
\sum_{n=1}^{\infty}2^{-3n} = \sum_{n=1}^{\infty}(2^{-3})^n = \frac{2^{-3}}{1 - 2^{-3}} = \frac{1/8}{1 - 1/8} = \frac{1/8}{7/8} = \frac{1}{7}.
$$
A: $$\sum_{n=0}^{+\infty}2^{-3n}=\sum_{n=0}^{+\infty}(2^{-3})^n=\sum_{n=0}^{+\infty}(\frac{1}{8})^n=\frac{1}{1-\frac{1}{8}}=\frac{1}{\frac{7}{8}}=\frac{8}{7}$$
A: Let $A=\sum_{n=1}^{\infty} 2^{-3n}$.  Then $8A = \sum_{n=1}^{\infty} 2^{-3(n-1)}= \sum_{n=0}^{\infty} 2^{-3n}.$ Then $$7A = 8A-A = \sum_{n=0}^{\infty} 2^{-3n} -\sum_{n=1}^{\infty} 2^{-3n} = 2^{-3\cdot0} = 1.$$  So $A=1/7$.
A: 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} $
