Another interesting fraction

[soroban]

$\displaystyle \text{We have: }\:\dfrac{1}{89} \;=\;0.01123595\,\,.\,.\,.$


$\displaystyle \text{The decimal is formed like this:}$

. . $\displaystyle 0.0{\color{blue}1}$
. . $\displaystyle 0.00{\color{blue}1}$
. . $\displaystyle 0.000{\color{blue}2}$
. . $\displaystyle 0.0000{\color{blue}3}$
. . $\displaystyle 0.00000{\color{blue}5}$
. . $\displaystyle 0.000000{\color{blue}8}$
. . $\displaystyle 0.000000{\color{blue}{13}}$
. . $\displaystyle 0.0000000{\color{blue}{21}}$
. . $\displaystyle 0.00000000{\color{blue}{34}}$
. . . . . . $\displaystyle \vdots$


$\displaystyle \displaystyle\text{It seems that: }\:\frac{1}{10}\sum^{\infty}_{n=1} \frac{F_n}{10^n} \;=\;\frac{1}{89}$

. . $\displaystyle \text{where }F_n\text{ is the }n^{th}\text{ Fibonacci number.}$


$\displaystyle \text{Care to prove it?}$