What is the sum of 0.1+ 0.01+ 0.002+ 0.0003+ 0.00005+ 0.000008+ ..., where each term is the n-th Fibonacci number, shifted n places to the right (that is, divided by 10^n)?
There is a famous formula that gives the n-th Fibonacci number: ((½(1+√5))^n -(½(1-√5))^n)/√5.
We
are then asked for the sum, for n=0 to infinity, of
f(n)/10^n.
This equals the sum of two geometric series,
so applying the usual formula we get (1/(1-(1+√5)/20) -
1/(1-(1+√5)/20))/√5, which works out to be 10/89.
Edited on March 9, 2005, 9:19 pm
Explained solution
