# Nested radical eventually reaches 1

Can anyone prove / disprove$$\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{...\sqrt{\sqrt{\sqrt{n}}}}}}}}} = 1$$ for all $$n>0$$?

$$\infty \sqrt{ }s$$

Note: This is my first question here. Any constructive feedback is welcomed. If you are wondering, this is not a piece of homework, but a problem I encountered when trying to solve a problem in Puzzling.SE. Thanks in advance!

• Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer. – dantopa Jun 9 '19 at 4:25
• You need to be more precise here. There's no accepted definition for applying a function infinitely many times. You could instead define $a_0 = n$ and $a_m = \sqrt{a_{m-1}}$ for all $m \ge 1$ (i.e. $a_m$ is the result of applying the square root function $m$ times to $n$), and then compute the limit. This limit will indeed equal $1$. – Theo Bendit Jun 9 '19 at 4:26
• @TheoBendit sorry i don't get what you mean. thanks! – Omega Krypton Jun 9 '19 at 4:31

Given that, the answer is indeed $$1$$ for any $$n>0$$. If you iterate taking the square roots $$m$$ times starting from $$n$$, that gives you $$n^{1/2^m}$$. As $$m\to\infty$$, the exponent converges to $$0$$, and so since the function $$x\mapsto n^x$$ is continuous, $$n^{1/2^m}$$ converges to $$n^0=1$$.