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)?

Let X = .1 + .01 + .002 + .0003 + .00005 + .000008 + .0000013 ...
Then 10X= 1 + .1 + .02 + .003 + .0005 + .00008 + .000013 + .0000021..
& 10X-1= .1 + .02 + .003 + .0005 + .00008 + .000013 + .0000021..
But this can be separated into two series:

.1 + .01 + .002 + .0003 + .00005 + .000008 + .0000013 ...
+ .01 + .001 + .0002 + .00003 + .000005 + .0000008
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10X-1= .1 + .02 + .003 + .0005 + .00008 + .000013 + .0000021..

We recognize the top series as X and the bottom series as X/10.
So 10X-1 = X + X/10
Solving for X gives us the solution: 10/89

Posted by Eric on 2005-03-10 06:40:39