# Is there a reason why $\frac{1}{7}$ as a decimal so perfectly seems to follow multiples of $7$?

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

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?

• This follows from the geometric series $1/49=0.02/(1-0.02)=\sum_{n\ge1}0.02^n$. Mar 27, 2023 at 17:33
• Yep. Big coincidence. Mar 27, 2023 at 17:34
• 1/7 is actually a cyclic number Apr 1, 2023 at 12:30
• @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. Apr 6, 2023 at 16:30
• Well, yeah, that's why I left as comment :P @WillOrrick Apr 6, 2023 at 22:29

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

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