Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?
share|improve this question
    
There is no rational $A_n$ other than $A_1$. –  Thomas Andrews Apr 29 '14 at 13:32
    

2 Answers 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 $k=1$).

share|improve this answer
    
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... –  ArtB Apr 29 '14 at 16:47
7  
@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. –  Henning Makholm 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 <q>If there were an irrational number with a rational square root,</q> –  ArtB Apr 30 '14 at 0:03

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

share|improve this answer
    
Very Nice!!!!!! –  barak manos Apr 29 '14 at 13:48
    
And what about $\lim\limits_{n\to\infty}A_n$ itself? Has it been proved irrational? –  barak manos Apr 29 '14 at 13:51
    
I answered the first question only, @barakmanos.I have no idea about the second. –  Thomas Andrews Apr 29 '14 at 14:28

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.