Is the Nested Radical Constant rational or irrational?

Question

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?

Answer 1

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 $n>1$, $$(\sqrt n)^2<n-1+\sqrt{n}<(\sqrt{n}+1)^2$$

Answer 2

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 $k=1$).