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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!

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  • $\begingroup$ 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. $\endgroup$ – dantopa Jun 9 at 4:25
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    $\begingroup$ 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 $1$. $\endgroup$ – Theo Bendit Jun 9 at 4:26
  • $\begingroup$ @TheoBendit sorry i don't get what you mean. thanks! $\endgroup$ – Omega Krypton Jun 9 at 4:31
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I assume that by such an expression with "infinitely many radicals" you really mean the limit as the number of radicals goes to infinity. (But, you really really shouldn't use notation like this without being precise about what it means.)

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 $n^0=1$.

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  • $\begingroup$ thanks! that helped a lot $\endgroup$ – Omega Krypton Jun 9 at 4:33

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