Another interesting fraction

**Soroban** · May 28th 2012, 07:58 AM

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

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

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

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

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

$\text{Care to prove it?}$

**BobP** · May 29th 2012, 01:56 AM

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

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

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

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

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

$S=\frac{1}{89}.$

**Soroban** · May 30th 2012, 07:09 AM

Elegant proof, BobP !

**richard1234** · June 13th 2012, 11:56 PM

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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