A $\sqrt{2}$ is at the end (?!) of this infinite nested radical expression. Is that enough to show that $\pi$ is irrational? Also, I'm curious if there is a better way to write this? Thanks! $$\lim_{n\rightarrow\infty}\left[2^{n+5}\cdot\sqrt{\frac{1-\sqrt{\frac{\text{n times}...\sqrt{\frac{1+\sqrt{\frac{\sqrt{1+\frac{\sqrt{2}}{2}}}{2}}}{2}}}{2}}}{2}}\right]=\pi$$


1 Answer 1


No, that's not enough. There's a $\sqrt{2}$ at the end of $$\lim_{n\rightarrow\infty}\underbrace{\sqrt{\sqrt{\ldots\sqrt{2}}}}_{n \text{ square roots}}$$ as well. This doesn't stop the limit from being equal to $1$. Limits don't play nicely with ideas of irrationality and rationality, because both rational and irrational numbers are dense, meaning that every open interval contains at least one of each - but limits only specify things in terms of open intervals, so they're pretty much useless for determining rationality or irrationality*. This, of course, means that determining the irrationality of $\pi$ is rather difficult, because when you need it in analysis, it's usually defined from a limit.

It is also perhaps worth noting that showing irrationality of each term is not necessarily as trivial as seeing a $\sqrt{2}$ somewhere; for instance, as a contrived example, we have $$\sqrt{11+6\sqrt{2}} + \sqrt{6-4\sqrt{2}} = 5$$ where I've just chosen each larger square root to be the square of a number of the form $a+b\sqrt{2}$ and chosen the $b\sqrt{2}$ terms in each to cancel. However, each term of your expression really is irrational - you can prove that from three lemmas:

If $x$ is irrational and $a$ is rational, then $a+x$ is irrational.

If $x$ is irrational and $a$ is a non-zero rational, then $a\cdot x$ is irrational.

If $x$ is irrational then $\sqrt{x}$ is irrational.

None of these are too difficult to prove - and repeatedly applything them does give that each term of your limit is irrational - however you have to be careful, because these lemmas only suffice to give irrationality of a fairly small class of values - generally, you have to be much more careful than thinking that a single irrational value prevents a whole expression in which it appears from being rational. And, of course, this digression is solely about the terms of the limit - it has little bearing on whether $\pi$ itself is irrational or not.

(*There's perhaps an exception if you know that the limit converges really fast and stays away from rationals with low denominator - for instance, it's possible to prove that $e=\sum_{n=0}^{\infty}\frac{1}{n!}$ by an argument of this nature - but that's way more structure than a limit and has nothing to do with whether the partial sums were rational - indeed, they are all rational here, despite the limit being irrational).

  • $\begingroup$ Very nice answer. It helps clarify various myths commonly harbored by students. +1 $\endgroup$
    – Paramanand Singh
    Apr 27, 2020 at 10:45

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