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Math Help - Another interesting fraction

  1. #1
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    Another interesting fraction

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


    \text{The decimal is formed like this:}

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


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

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


    \text{Care to prove it?}
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  2. #2
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    Re: Another interesting fraction

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

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

    =\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}}

    =\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}  }

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

    S=\frac{1}{89}.
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  3. #3
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    Re: Another interesting fraction


    Elegant proof, BobP !

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  4. #4
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    Re: Another interesting fraction

    I like this one:

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


    It's very easy to prove. Let

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


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

    Solving for S, we obtain S = \frac{1000}{998999}. According to WolframAlpha, this decimal repeats with period 496,620.
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