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 May 28th, 2012, 08:09 AM #1 Math Team   Joined: Dec 2006 From: Lexington, MA Posts: 3,267 Thanks: 408 Another interesting fraction $\text{We have: }\:\dfrac{1}{89} \;=\;0.01123595\,\,.\,.\,.$ $\text{The decimal is formed like this:}$ [color=beige]. . [/color]$0.01$ [color=beige]. . [/color]$0.001$ [color=beige]. . [/color]$0.0002$ [color=beige]. . [/color]$0.00003$ [color=beige]. . [/color]$0.000005$ [color=beige]. . [/color]$0.0000008$ [color=beige]. . [/color]$0.00000013$ [color=beige]. . [/color]$0.000000021$ [color=beige]. . [/color]$0.0000000034$ [color=beige]. . . . . . [/color]$\vdots$ $\displaystyle\text{It seems that: }\:\frac{1}{10}\sum^{\infty}_{n=1} \frac{F_n}{10^n} \;=\;\frac{1}{89}$ [color=beige]. . [/color]$\text{where }F_n\text{ is the }n^{th}\text{ Fibonacci number.}$ $\text{Care to prove it?}$
 May 28th, 2012, 11:38 AM #2 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Re: Another interesting fraction It's a geometric series due to the Binet formula. I don't have pencil or paper at the moment but that should provide the answer readily enough.
 May 31st, 2012, 04:49 AM #3 Math Team   Joined: Apr 2012 Posts: 1,579 Thanks: 22 Re: Another interesting fraction I was hoping that this thread would get more traction. My own quickie internet search came up with the fact that 89 is of the form x^2-x-1 for the x that serves as the base of our decimal system, ie 10. I've had a general interest in the difference between results that are dependent on the base you are working in and those that are not. Divisibility tests are obviously totally base dependent. Want to know whether a large number is divisible by 137? Just translate the base ten numeral into base 137 and see if it ends in zero! Hey, thanks for the tip, eh? This one, however, seems more interesting.
May 31st, 2012, 07:41 AM   #4
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Re: Another interesting fraction

Quote:
 Originally Posted by johnr I was hoping that this thread would get more traction.
OK.

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So F_n = (phi^n + phi^-n)/sqrt(5) so
$\frac{1}{10}\sum_{n=1}^\infty\frac{F_n}{10^n}=\fra c{1}{10\sqrt5}\sum_{n=1}^\infty\left(\frac{\varphi ^n}{10^n}+\frac{\varphi^{-n}}{10^n}\right)=\frac{1}{10\sqrt5}\left(\sum_{n=1 }^\infty(\varphi/10)^n+\sum_{n=1}^\infty\frac{1}{(10\varphi)^n}\rig ht)=\frac{1}{10\sqrt5}\left(\frac{1}{10/\varphi-1}+\frac{1}{10\varphi-1}\right)$

This can be simplified mechanically (replace phi with radical definition, rationalize denominators three times, multiply through rationalize denominators again, add, reduce denominators by gcd). I get $\frac{2475+149\sqrt5}{242525}$ which is close to but not equal to 1/89.

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Oops, got the Binet formula wrong, should be F_n = (phi^n - (-phi)^-n)/sqrt(5). That does give 1/89. The proof is tedious and essentially identical to the above with some signs reversed.

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