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I am curious to know if $N=0.12233344444455555...$ is a rational or an irrational number. I see that, since it can be obtained by the sum of $0+0.1+0.022+0.000333+...$, it could be obtained by this infinite series:

$$N=\sum_{n=1}^\infty n\left(\sum_{m=1}^n 10^{m-1}\right)10^{-{n(n+1)}/{2}}$$

In addition, since it can also be obtained by the sum of $0.\overline{1}+0.0\overline{1}+0.000\overline{1}+...=\frac{1}{9}+\frac{1}{90}+\frac{1}{9000}+...$, it can also be obtained by a sum of infinite recurring decimals:

$$N=\sum_{n=0}^\infty \frac{1}{9·10^{{n(n+1)}/{2}}} =\frac{1}{9} \sum_{n=0}^\infty 10^{-{n(n+1)}/{2}}$$

Would this number, obtained as an infinite sum of recurring decimals, be also a recurring decimal? If not, since it is also the sum of infinite rational numbers, would it even be rational?

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    $\begingroup$ $N$ seems to be irrational (an infinite sum of rational numbers need not be rational at all). Maybe someone can at least prove that the decimal expansion of $N$ cannot terminate. Irrationality proofs are usually very diffucult, mabe Roth's theorem helps. $\endgroup$
    – Peter
    Dec 12, 2021 at 13:35
  • $\begingroup$ Analyzing the continued fraction of $N$ with PARI/GP, I found out that if $N$ is actually rational, numerator and denominator (even if written in the lowest terms) must have more than $4\ 000$ digits. $\endgroup$
    – Peter
    Dec 12, 2021 at 13:36
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    $\begingroup$ The convergence is so fast that the number could even be a Liouville number, but even if not, Roth's improvement could allow a transcendentality proof, if we actually can prove the irrationality. I checked $N$ with the algdep.command with higher degrees and $N$ seems even to be transcendental. $\endgroup$
    – Peter
    Dec 12, 2021 at 13:45
  • $\begingroup$ The second way Invenietis has written the sum makes it clear that the number $N$ is irrational; the decimal expansion of $9N$ consists of zeros and ones with its nonzero entries spaced ever further apart. $\endgroup$
    – hardmath
    Jan 3, 2022 at 3:35
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    $\begingroup$ Can't you do that with any number? $$\pi=3+0.1+0.04+0.001+0.0005+0.00009+0.000002+\cdots$$ $\endgroup$
    – bof
    Jan 3, 2022 at 3:49

1 Answer 1

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The second way you've written the summation makes it clear the number $N$ is irrational, because $9N$ is non-terminating and non-recurring:

$$ 9N = 1.10100100001000001\ldots $$

Note that location of the digit $1$ from the $10^{-(n-1)n/2}$ term is separated from the next corresponding digit $1$ from the $10^{-n(n+1)/2}$ term by $n$ decimal places, i.e. by $n-1$ intervening zero digits. Thus the decimal expansion never becomes repeating of any finite length digit pattern. But any rational number would so repeat, and $N$ is rational if and only if $9N$ is rational.

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