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Does Champernowne's constant converge to the digits of $\pi$?

Clearly if you go far enough into Champernowne's constant you will find a string of the first $n$ digits of $\pi$ for any finite $n$.

For every finite set of $n$ digits of $\pi$ contained within the constant, $n+1$ digits are subsequently represented $a(n+1)$ digits later, for some $a\leq b$ where $b$ is the base.

By induction there is no string of digits which is not, ultimately, represented.

If it contains the digits of $\pi$, it cannot contain anything after them since they never end, so if it contains them, it must contain them at its end. Could it converge to the digits of $\pi$?

However it would seem we can never reach them if it did, since they would occur at a length greater than the length of the digits of $\pi$. Does this mean they are not represented?

it's a hire car baby
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  • Just because any finite string of digits of $\pi$ is represented, doesn't mean that the infinite expansion of $\pi$ is in there. There simply isn't enough room. – Arthur Apr 07 '17 at 17:12
  • @Arthur there may not be enough room. That is the question I guess. Pi is pretty long, but I think the room in there is pretty limitless also. I'm looking for hard logic or reason I guess. Or at least argument. – it's a hire car baby Apr 07 '17 at 17:13

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Here's a heuristic that points out the flaw in this reasoning: Exactly the same logic works for any other non-repeating non-terminating decimal. So consider the representation of $\sqrt{2}$. It can't 'converge' to both, and in fact I think you'll find this convergence notion doesn't really hold much water if you examine it closely. In particular, you're trying to say something like, this constant times some big power of 10 is equal to a huge number + $\pi$. But what would this huge power of $10$ be when you have to keep going out further and further to get longer strings of digits? The sequence you would use to approximate, which you would then take the limit of, would have to diverge.

A. Thomas Yerger
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  • Ok that kind of holds water. But why does that (divergence) not apply to Champernowne's constant itself? I'm not saying it doesn't I just can't see why just now. – it's a hire car baby Apr 07 '17 at 18:12
  • Because it's a sum of the form $\sum \frac{a_i}{10_i}$, where the $a_i$ are digits. This sum converges when all that $a_i$ are $9$, and so this is a monotonic sequence of partial sums, which is bounded above. Thus it converges. – A. Thomas Yerger Apr 07 '17 at 18:32