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$\frac{1}{7} = 0.\overline{142857}$

So it's easy to learn. Double $7$ three times, then add $1$ to the end to set up for a repeat.

$0.(2 \times 7 = 14)(2 \times 14 = 28)(2 \times 28 = 56 + 1 = 57)$, $+1$ indicates it's time to repeat because we've reached $7$ and can start doubling again to get $14$.

Is this just a big coincidence?

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    $\begingroup$ This follows from the geometric series $1/49=0.02/(1-0.02)=\sum_{n\ge1}0.02^n$. $\endgroup$
    – Kenta S
    Mar 27, 2023 at 17:33
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    $\begingroup$ Yep. Big coincidence. $\endgroup$ Mar 27, 2023 at 17:34
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    $\begingroup$ 1/7 is actually a cyclic number $\endgroup$ Apr 1, 2023 at 12:30
  • $\begingroup$ @TrystwithFreedom I guess it's $142857$ that's the cyclic number? At least that's what I gather from the Wikipedia article. I seems that $142857$ being cyclic follows from $7$ being prime and having digital period $6=7-1$. Although this definitely seems related to the question asked here, I don't see that it fully answers it. $\endgroup$ Apr 6, 2023 at 16:30
  • $\begingroup$ Well, yeah, that's why I left as comment :P @WillOrrick $\endgroup$ Apr 6, 2023 at 22:29

1 Answer 1

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Hidden in the expression $$\dfrac 17 = 0.142857142857\cdots$$ is a geometric sequence.

The quick way of looking at this is to see that $142857$ can be broken up into $14$, $28$ and $57$, which is very close to $14$, $28$, and $56$; a geometric progression with a common ratio of $2$. The $1$ that would make $56$ into $57$ could come from the first $1$ of $112$ which would be the next number of the sequence $14$, $28$, $56$.

Here is the long way. Note that, starting with $0.14$, each row is $\dfrac{2}{100}$ times the previous row. If you add up the terms below the horizontal line, you should get $\dfrac 17$.

$$\begin{array}{rrrrrrrrrrr} 0&. & 14& 28&57 &14 & 28& 57&\cdots & = &\dfrac 17\\ \hline \\ 0&. & 14& & & & & &\cdots & =& 1 \cdot \dfrac{14}{10^ 2}\\[12pt] 0&. & 00& 28& & & & &\cdots & =& 2 \cdot \dfrac{14}{10^ 4}\\[12pt] 0&. & 00& 00& 56& & & &\cdots & =& 4 \cdot \dfrac{14}{10^ 6}\\[12pt] 0&. & 00& 00& 01& 12& & &\cdots & =& 8 \cdot \dfrac{14}{10^ {8}}\\[12pt] 0&. & 00& 00& 00& 02& 24& &\cdots & =& 16 \cdot \dfrac{14}{10^{10}}\\[12pt] 0&. & 00& 00& 00& 00& 04& 48&\cdots & =& 32 \cdot \dfrac{14}{10^{12}}\\[12pt] 0&. & 00& 00& 00& 00& 00& 08&\cdots & =& 64 \cdot \dfrac{14}{10^{14}}\\[12pt] \end{array}$$

This suggests that

$ \dfrac 17 = 1 \cdot \dfrac{14}{10^2} + 2 \cdot \dfrac{14}{10^4} + 4 \cdot \dfrac{14}{10^6} + 8 \cdot \dfrac{14}{10^8} + 16 \cdot \dfrac{14}{10^{10}} + 32 \cdot \dfrac{14}{10^{12}} + 64 \cdot \dfrac{14}{10^{14}} + \cdots$

The sequence on the right is a geometric sequence with an initial term of $t_0 = \dfrac{14}{100}$ and a common ratio of $r = \dfrac{2}{100}$. The sum is

$$ 0.\overline{142857} = \dfrac{t_0}{1-r} = \dfrac{ \left( \dfrac{14}{100} \right) }{\left( \dfrac{98}{100} \right)} =\dfrac 17 $$.

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