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
         ....
-------------
.12499999744
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We have that:

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

$\ds \sum_{k \mathop \ge 0} \dfrac {2^k} {10^{k + 1} }$

$=$

$\ds \dfrac {1/10} {1 - 1/5}$

$\ds $

$=$

$\ds \dfrac 1 {10 \paren {4 / 5} }$

$\ds $

$=$

$\ds \dfrac 5 {40}$

$\ds $

$=$

$\ds \dfrac 1 8$

$\blacksquare$