If one were to take the Fibonacci numbers and arrange them so that the units digit of the nth Fibonacci number is in the $(n+1)th$ decimal place and sum these numbers up they would get the reciprocal of the eleventh number $\frac{1}{89}$ or 0.011235955. This might look something like:
0.01 + .0001 + 0.0002 + 0.00003 + 0.000005+ 0.0000008 + 0.00000013 + 0.000000021 + 0.0000000034 + 0.00000000055
= 0.011235955
I am trying to prove this but I am getting turned around. So far I have the summation where $$S = \sum_{n=0}^{\infty} \frac{F_n}{10^{n+1}}$$ $$ = \frac{0}{10^{1}} + \frac{1}{10^2} + \frac{1}{10^3} + \frac{2}{10^4} ... +$$
but I am unsure of where to go from here. Can anyone please provide some insight? Thanks.
