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Show that the sequence $\{a_n\}_{n\in \mathbb{N}} = \{x: x=$ the nth decimal digit of Champernowne's constant$\}$ is not periodic.

For those who don't know what Champernowne's constant is, it's the following number:

$C_{10} = 0.123456789101112131415161718192021...$

Well, before someone says that $\{a_n\}_{n\in \mathbb{N}}$ is definitely aperiodic because Champernowne's constant is irrational(and even transcendental), I have to say that I'm well aware of the fact that a number is rational if and only if the sequence of its digits is periodic, but I'm looking for another proof that the sequence of the digits of $C_{10}$ is aperiodic that doesn't use anything but a smart way of showing that $a_{n+k} = a_n$ can't hold for any $k\in \mathbb{N}$.

I've observed that by grouping the terms of a sequence in a clever way we can show that no such $k$ can exist for some sequences. For example I created the following sequence:

$1,0,0,1,1,1,0,0,0,0,1,1,1,1,1,0,0,0,0,0,0,1,1,1,1,1,1,1,\cdots$

The sequence is created this way: First you put one $1$, then two $0$'s, and in the nth step you add either $1$'s or $0$'s n times depending on whether n is odd or even respectively.

It's easy to show this sequence is not periodic, because no matter how large $k$ is you can always find $n$ such that $a_n = 1$ but $a_{n+k}=0$ or $a_n=0$ but $a_{n+k}=1$. That proves the sequence is not periodic and as a corollary it' proved that the real number $0.1001110000111110000001111111\cdots$ is irrational.

After showing that I've been trying to apply this technique of grouping to the sequence in problem without much success so far.

I think if the periodicity of a sequence $\{a_n\}$ is denoted by $\Omega(\{a_n\})$ then if we define a new sequence $\bar{a}_m(n) := \bar{a}_{m,n} = [a_n]_m$ then $\Omega(\{\bar{a}_{m,n}\}) \mid \Omega({a_n})$. This shouldn't be very hard to prove I think.

Now, to show that

$1,2,3,4,5,6,7,8,9,1,0,1,1,1,2,1,3,1,4,1,5,1,6,1,7,1,8,1,9,2,0,2,1,\cdots$

is not periodic it's sufficient to prove that the sequence

$1,0,1,0,1,0,1,0,1,1,0,1,1,1,0,1,0,1,0,1,\cdots$

is not periodic. But I haven't progressed much from here :/

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2 Answers 2

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$C_{10}$ contains arbitrarily long sequences of zeroes, but is not zero.

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  • $\begingroup$ Doesn't that observation need a proof? $\endgroup$
    – user66733
    Aug 30, 2013 at 17:09
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    $\begingroup$ @some1: No, it is a proof in itself. A simple and beautiful proof. $\endgroup$
    – TonyK
    Aug 30, 2013 at 17:12
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    $\begingroup$ @some1.new4u At some point in the decimal expansion of Champernowne's constant, the number $10^k$ is used for the digits. $\endgroup$
    – Daniel Fischer
    Aug 30, 2013 at 17:12
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    $\begingroup$ Here's an elaboration. As Daniel said, $C_{10}$ contains arbitrarily long sequences of zeroes, because for every $k$, it contains $10^k$ in its decimal expansion. Now suppose $a_n$ is periodic with period $p$. But we can find $p$ zeroes in a row in the sequence, which means the entire period is just zeroes. Therefore, the entire sequence must be 0. Contradiction! $\endgroup$
    – Lopsy
    Aug 30, 2013 at 17:14
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    $\begingroup$ Ah, now you're spoiling it Lopsy! $\endgroup$
    – TonyK
    Aug 30, 2013 at 17:14
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Less elegant, but more direct to my taste.

Let's say there is some period. At some point after the period start you'll get a sequence beginning with that period and followed by the first digit of the period plus 1 modulo 10. It's true because any given sequence appears an infinite number of times in Champernowne's constant. Which contradicts the possibility of a period.

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