# 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 $10$:

$\dfrac 1 8 = \ds \sum_{k \mathop \ge 0} \dfrac {2^k} {10^{k + 1} }$
.1
2
4
8
16
32
64
128
256
512
1024
....
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.12499999744
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## Proof

We have that:

$\dfrac {2^k} {10^{k + 1} } = \dfrac 1 {10} \paren {\dfrac 1 5}^k$

and so by Sum of Infinite Geometric Sequence:

 $\ds \sum_{k \mathop \ge 0} \dfrac {2^k} {10^{k + 1} }$ $=$ $\ds \dfrac {1/10} {1 - 1/5}$ $\ds$ $=$ $\ds \dfrac 1 {10 \paren {4 / 5} }$ $\ds$ $=$ $\ds \dfrac 5 {40}$ $\ds$ $=$ $\ds \dfrac 1 8$

$\blacksquare$