Reciprocal of 89 as Sum of Fibonacci Numbers by Negative Powers of 10

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Theorem

$\ds \sum_{k \mathop \ge 0} \dfrac {F_k} {10^{k + 1} } = \dfrac 1 {89}$

where $F_k$ is the $k$th Fibonacci number:

$F_0 = 0, F_1 = 1, F_k = F_{k - 1} + F_{k - 2}$


That is: 

1 / 89 = 0.0
       + 0.01
       + 0.001
       + 0.0002
       + 0.00003
       + 0.000005
       + 0.0000008
       + 0.00000013
       + 0.000000021
       + 0.0000000034
       + 0.00000000055
       + ..............

Proof

First we note that from Reciprocal of $89$:

$\dfrac 1 {89} = 0 \cdotp \dot 01123 \, 59550 \, 56179 \, 77528 \, 08988 \, 76404 \, 49438 \, 20224 \, 719 \dot 1$


We have that:

$89 = 10^2 - 10 - 1$

So:

\(\ds \sum_{k \mathop \ge 0} \dfrac {F_k} {10^{k + 1} }\) \(=\) \(\ds \dfrac 1 {10} \sum_{k \mathop \ge 0} F_k \paren {\dfrac 1 {10} }^k\)
\(\ds \) \(=\) \(\ds \dfrac 1 {10} \cdot \dfrac {\dfrac 1 {10} } {1 - \dfrac 1 {10} - \paren {\dfrac 1 {10} }^2 }\) Generating Function for Fibonacci Numbers
\(\ds \) \(=\) \(\ds \dfrac 1 {10^2 - 10 - 1}\)
\(\ds \) \(=\) \(\ds \dfrac 1 {89}\)

$\blacksquare$


Sources

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