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Given the sequence $A_n=\sqrt{1+\sqrt{2+\sqrt{3+\sqrt{\dots+\sqrt{n}}}}}$:

  • Are there any known rational elements in $A_n$, or has it been proved that all are irrational?
  • Is there any proof for $\lim\limits_{n\to\infty}A_n$ (a.k.a. Nested Radical Constant) to be rational or irrational?
barak manos
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2 Answers2

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The square of a rational number is always rational, so the square root of an irrational is also necessarily irrational.

Therefore, as you go through the sequence $$\sqrt n, \sqrt{(n-1)+\sqrt{n}}, \sqrt{(n-2)+\sqrt{(n-1)+\sqrt{n}}}, \ldots, A_n$$ once you hit even one irrational number, $A_n$ at the far end of the sequence will be irrational too.

If $n$ is not a perfect square, this makes $A_n$ irrational right away.

On the other hand, if $n$ is a perfect square, $n=k^2$, then the second element in the computation is $\sqrt{k^2+k-1}$, which is strictly between $k$ and $k+1$ and therefore irrational (unless $k=1$).

hmakholm left over Monica
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    Gotcha, thanks! And what about $\lim\limits_{n\to\infty}A_n$ itself? Has it been proved irrational? – barak manos Apr 29 '14 at 13:50
  • How does that follow? The square of a rational number is always rational, so the square root of an irrational is also necessarily irrational. The first part seems obvious to me, the second less so... – Sled Apr 29 '14 at 16:47
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    @ArtB: If there were an irrational number a rational square root, then the square of the rational square root would be the original irrational number, which contradicts the fact that the square of a rational number is rational. – hmakholm left over Monica Apr 29 '14 at 17:19
  • Yeah, that makes sense once you spell it out like that. Thanks for that! To anyone in the future, I believe he meant to say If there were an irrational number with a rational square root, – Sled Apr 30 '14 at 00:03
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For the first question: If $A_n$ is rational, we can prove that $\sqrt{n}$ is rational, thus $n$ is a perfect square, and then we can prove that $\sqrt{n-1+\sqrt{n}}$ is rational, and hence $n-1+\sqrt n$ is a perfect square. But if $n>1$, $$(\sqrt n)^2<n-1+\sqrt{n}<(\sqrt{n}+1)^2$$

Thomas Andrews
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