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

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: