the scary fraction of one seventh

[MattcAnderson]

Okay, it is not that scary.

Some of us know that 1/7 can be represented as an infinite repeating decimal.
The six digits are 1,4,2,8,5, and7
So here it is with those digits twice, because we all want to see it twice :)

1/7 = 0.142857 142857

Regards,
Matt

[Dobri]

Here is a link to more fun facts about 142857, see https://en.wikipedia.org/wiki/142,857.

[kruoli]

Edit: This first part is also seen on the page Dobri provided. I was slower with my writeup than he with a link. The arithmethic progression shown there based on 3 is also interesting.

You can also see how \[\frac{1}{7}=\sum^{\infty}_{n=1}{\frac{2^n\cdot{}7}{10^{2n}}}\]
by "visually inspecting" the terms:

x | value of x-th term
--------------
1 | 0.1400000000
2 | 0.0028000000
3 | 0.0000560000
4 | 0.0000011200
5 | 0.0000000224
 ...
-----------------
  = 0.142857142...

Of course this is not a proof, this is only a neat visualisation, similar to the one for \[\frac{1}{89}=\sum^{\infty}_{n=0}{\frac{F_n}{10^{n+1}}}\] where \(F_n, n \in \mathbb{N}_0\) are the Fibonacci numbers.

This latter one can be generalised: For \(\frac{1}{89}\), we have "one digit per \(F_n\)". If you want to have \(d\) digits, you should take \[\frac{1}{10^{2d}-10^d-1}=\sum^{\infty}_{n=0}{\frac{F_n}{10^{d\cdot{}n+1}}}\] instead, e.g. \(d=4\): \[\frac{1}{10^{2d}-10^d-1}=\frac{1}{10^8-10^4-1}=\frac{1}{99989999}=0.\ 0000\ 0001\ 0001\ 0002\ 0003\ 0005\ 0008\ 0013\ 0021\ 0034\ 0055\ 0089\ 0144 \dots\]

[MattcAnderson]

The two previous posts are very cool.