Reciprocal of 8

Theorem
The reciprocal of $8$ can be expressed as the summation of the powers of $2$ multiplied by the reciprocal powers of $10$:
 * $\dfrac 1 8 = \displaystyle \sum_{k \mathop \ge 0} \dfrac {2^k} {10^{k + 1} }$

.1 2   4    8    16     32      64      128       256        512        1024         .... - .12499999744 -

Proof
We have that:
 * $\dfrac {2^k} {10^{k + 1} } = \dfrac 1 {10} \left({\dfrac 1 5}\right)^k$

and so by Sum of Infinite Geometric Progression: