A series for Fibonacci numbers. How can I prove The Fibonacci sequence is encoded in the number $1/89$  i.e. $( 1/89 = 0.01 + 0.001 + 0.0002 + 0.00003 + 0.000005 + 0.0000008 + 0.00000013 + 0.000000021 + 0.0000000034 \ldots)$
 A: If there were such $x$ ($ = 0.011235\ldots$), then this satisfies the property of Fibonacci sequence:
$$\begin{array}{lllllllll}
&0.&1&1&2&3&5&\ldots&//\ 10x\\
+&0.&0&1&1&2&3&\ldots&//\ x\\
\hline
&0.&1&2&3&5&9&\ldots\\
+&1\\
\hline
&1.&1&2&3&5&9&\ldots&//\ 100x
\end{array}
$$
So by solving for such $x$,
$$\begin{align*}
10x + x &= 100x - 1\\
89x &= 1\\
x &= \frac{1}{89}\end{align*}$$
A: I like peterwhy’s answer — think of this as an extended comment on it — but it does slide over a potentially tricky point: $x$ isn’t
$$\begin{array}{lll}
&0.&0&1&1&2&3&5&\color{brown}8&\color{brown}13&\ldots\;,
\end{array}$$
as the picture might suggest, but rather
$$\begin{array}{lll}
&0.&0&1&1&2&3&5&\color{brown}9&5&\ldots\;,
\end{array}$$
because of carries. We actually have
$$x=\sum_{n\ge 1}\frac{F_{n-1}}{10^n}\;.$$
Now we can carry out peterwhy’s calculation as follows:
$$\begin{align*}
0.01+0.1x+0.01x&=0.01+\sum_{n\ge 1}\frac{F_{n-1}}{10^{n+1}}+\sum_{n\ge 1}\frac{F_{n-1}}{10^{n+2}}\\\\
&=0.01+\sum_{n\ge 2}\frac{F_{n-2}}{10^n}+\sum_{n\ge 3}\frac{F_{n-3}}{10^n}\\\\
&=0.01+\frac{F_0}{10^2}+\sum_{n\ge 3}\frac{F_{n-2}+F_{n-3}}{10^n}\\\\
&=0.01+\sum_{n\ge 3}\frac{F_{n-1}}{10^n}\\\\
&=\sum_{n\ge 1}\frac{F_{n-1}}{10^n}\\\\
&=x\;,
\end{align*}$$
so $0.01=0.89x$, and $x=\dfrac1{89}$.
