Results 1 to 4 of 4

Thread: Another interesting fraction

  1. #1
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    12,028
    Thanks
    849

    Another interesting fraction

    $\displaystyle \text{We have: }\:\dfrac{1}{89} \;=\;0.01123595\,\,.\,.\,.$


    $\displaystyle \text{The decimal is formed like this:}$

    . . $\displaystyle 0.0{\bf1}$
    . . $\displaystyle 0.00{\bf1}$
    . . $\displaystyle 0.000{\bf2}$
    . . $\displaystyle 0.0000{\bf3}$
    . . $\displaystyle 0.00000{\bf5}$
    . . $\displaystyle 0.000000{\bf8}$
    . . $\displaystyle 0.000000{\bf{13}}$
    . . $\displaystyle 0.0000000{\bf{21}}$
    . . $\displaystyle 0.00000000{\bf{34}}$
    . . . . . . $\displaystyle \vdots$


    $\displaystyle \displaystyle\text{It seems that: }\:\frac{1}{10}\sum^{\infty}_{n=1} \frac{F_n}{10^n} \;=\;\frac{1}{89}$

    . . $\displaystyle \text{where }F_n\text{ is the }n^{th}\text{ Fibonacci number.}$


    $\displaystyle \text{Care to prove it?}$
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Jun 2009
    Posts
    671
    Thanks
    136

    Re: Another interesting fraction

    Let $\displaystyle S=\frac{1}{10}\sum^{\infty}_{n=1}\frac{F_{n}}{10^{ n}},$ then

    $\displaystyle 10S=\frac{F_{1}}{10}+\sum^{\infty}_{n=2}\frac{F_{n +1}-F_{n-1}}{10^{n}}$

    $\displaystyle =\frac{1}{10}+10\sum^{\infty}_{n=2}\frac{F_{n+1}}{ 10^{n+1}}-\frac{1}{10}\sum^{\infty}_{n=2}\frac{F_{n-1}}{10^{n-1}}$

    $\displaystyle =\frac{1}{10}+10\sum^{\infty}_{n=1}\frac{F_{n}}{10 ^{n}}-10\{\frac{F_{2}}{10^{2}}+\frac{F_{1}}{10}\}}-\frac{1}{10}\sum^{\infty}_{n=1}\frac{F_{n}}{10^{n} }$

    $\displaystyle =\frac{1}{10}+100S-\frac{1}{10}-1-S$

    $\displaystyle S=\frac{1}{89}.$
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    12,028
    Thanks
    849

    Re: Another interesting fraction


    Elegant proof, BobP !

    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Jun 2012
    From
    AZ
    Posts
    616
    Thanks
    97

    Re: Another interesting fraction

    I like this one:

    $\displaystyle \frac{1000}{998999} = 0.001001002003005008013...$


    It's very easy to prove. Let

    $\displaystyle S = .001001002003005008... = \sum_{i=1}^{\infty} (\frac{1}{1000})^{i} F_i$
    $\displaystyle \frac{S}{1000} = .000001001002003005...$


    Adding them up, we get $\displaystyle \frac{1001S}{1000} = .001002003005008... = 1000S - 1$

    Solving for S, we obtain $\displaystyle S = \frac{1000}{998999}$. According to WolframAlpha, this decimal repeats with period 496,620.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. An interesting fraction
    Posted in the Math Puzzles Forum
    Replies: 3
    Last Post: Apr 25th 2012, 08:52 AM
  2. Simplifying fraction over fraction expression
    Posted in the Algebra Forum
    Replies: 3
    Last Post: Oct 17th 2010, 02:27 PM
  3. please help improper fraction into partial fraction
    Posted in the Pre-Calculus Forum
    Replies: 7
    Last Post: Mar 1st 2010, 09:06 AM
  4. Replies: 2
    Last Post: Sep 28th 2009, 07:18 PM
  5. Replies: 4
    Last Post: Aug 31st 2007, 11:05 PM

Search Tags


/mathhelpforum @mathhelpforum