\(\displaystyle \text{We have: }\:\dfrac{1}{89} \;=\;0.01123595\,\,.\,.\,.\)
\(\displaystyle \text{The decimal is formed like this:}\)
. . \(\displaystyle 0.0{\color{blue}1}\)
. . \(\displaystyle 0.00{\color{blue}1}\)
. . \(\displaystyle 0.000{\color{blue}2}\)
. . \(\displaystyle 0.0000{\color{blue}3}\)
. . \(\displaystyle 0.00000{\color{blue}5}\)
. . \(\displaystyle 0.000000{\color{blue}8}\)
. . \(\displaystyle 0.000000{\color{blue}{13}}\)
. . \(\displaystyle 0.0000000{\color{blue}{21}}\)
. . \(\displaystyle 0.00000000{\color{blue}{34}}\)
. . . . . . \(\displaystyle \vdots\)
\(\displaystyle \displaystyle\text{It seems that: }\:\frac{1}{10}\sum^{\infty}_{n=1} \frac{F_n}{10^n} \;=\;\frac{1}{89}\)
. . \(\displaystyle \text{where }F_n\text{ is the }n^{th}\text{ Fibonacci number.}\)
\(\displaystyle \text{Care to prove it?}\)