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I was thinking what happens with the sequence $\{x_n\}_{n\in \Bbb N}$ where:

$$x_{n}:= \sqrt[n]{n \sqrt[n]{n \sqrt[n]{n\ldots}}}$$

When you look some terms, for example $x_{1}=1$, $x_{2}=\sqrt[]{2 \sqrt[]{2 \sqrt[]{2 ...}}}$, $x_{3}=\sqrt[3]{3 \sqrt[3]{3 \sqrt[3]{3 ...}}}$, these terms and the others will be continued fractions, where each one converges.

I'm asking what happens with $\lim\limits_{n \to \infty}\sqrt[n]{n \sqrt[n]{n \sqrt[n]{n ...}}}$ ?. I have an idea and it is $\lim\limits_{n \to \infty}\sqrt[n]{n \sqrt[n]{n \sqrt[n]{n ...}}}=1$. My reasoning is in the fact:

$$\sqrt[n]{n \sqrt[n]{n \sqrt[n]{n ...}}}= \displaystyle {n^{\frac{1}{n}}} n^{\frac{1}{n^2}} n^{\frac{1}{n^3}}...$$

And you know that:

$${\displaystyle \frac{1}{n}> \frac{1}{n^k} \textrm{ for } n,k \in \Bbb N}$$

Then:

$$n^{\frac{1}{n}}> n^{\frac{1}{n^k}} \geq 1$$

Like $\lim\limits_{n \to \infty}n^{\frac{1}{n}}=1$ and $\lim\limits_{n \to \infty}1=1$, by the Squeeze Theorem $\lim\limits_{n \to \infty}\sqrt[n]{n \sqrt[n]{n \sqrt[n]{n ...}}}=1$. Is this reasoning correct? What do you think about $x_{n}$? Do you think there is another way to prove it? I receive suggestions or comments. Thank you.

Asaf Karagila
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angie duque
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  • You have not actually defined $x_n$, so much as described things about it (using the $\ldots$ symbol) that you want to be true. Until you have a definition of $x_n$, there is no "it" to prove. – kimchi lover Jul 03 '20 at 14:21
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    @kimchi, I think this kind of infinite nested radical is common enough not to need defining every time it's used. You know what it means. – TonyK Jul 03 '20 at 14:32
  • @TonyK The "continuos fraction" $\sqrt[n]{n\sqrt[n]{\dots}}$ is not familiar to me. As a a general matter of math hygene, before one uses the values of limits in definitions, one should prove they exist. Would it have hurt the OP to define $x_n=n^{1/(n-1)}$ and then note that $x_n=\sqrt[n]{nx_n}$, and so on? – kimchi lover Jul 03 '20 at 14:38
  • This answer might be helpful as well as this question. – PinkyWay Jul 03 '20 at 15:15
  • These are not fractions! – Ivan Neretin Jul 03 '20 at 19:05
  • Some relevant posts you can find: https://approach0.xyz/search/?q=%24%5Csqrt%5Bn%5D%7Bn%20%5Csqrt%5Bn%5D%7Bn%20%5Csqrt%5Bn%5D%7Bn%5Cldots%7D%7D%7D%24&p=1 – Wei Zhong Jul 04 '20 at 06:36

5 Answers5

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Yes, the limit is $1$: $$x_n=n^{\sum_{k=1}^{\infty}\frac{1}{n^k}}= n^{\frac{1}{n-1}}=e^{\frac{\ln(n)}{n-1}}\to 1.$$

Robert Z
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An Inductive idea is: $$x_{2}=\sqrt[]{2 \sqrt[]{2 \sqrt[]{2 ...}}}\to 2\\ x_{3}=\sqrt[3]{3 \sqrt[3]{3 \sqrt[3]{3 ...}}}\to \sqrt[2]3\\ x_{4}=\sqrt[4]{4 \sqrt[4]{4 \sqrt[4]{4 ...}}}\to \sqrt[3]4\\ x_{5}=\sqrt[5]{5 \sqrt[5]{5 \sqrt[5]{5 ...}}}\to \sqrt[4]5\\\vdots\\ x_{n}= \sqrt[n]{n \sqrt[n]{n \sqrt[n]{n ...}}}\to \sqrt[n-1]n$$and it tends to $$\sqrt[n-1]n=n^{\frac{1}{n-1}}\to 1$$
Implicit : idea to solve for example$$\sqrt[3]{3 \sqrt[3]{3 \sqrt[3]{3 ...}}}=a\to \text{to the power of 3}\\a^3=3\sqrt[3]{3 \sqrt[3]{3 \sqrt[3]{3 ...}}}\\a^3=3\underbrace{\sqrt[3]{3 \sqrt[3]{3 \sqrt[3]{3 ...}}}}_{a}\\a^3=3a\underbrace{\to}_{a\neq 0}a^2=3\to a=\sqrt 3$$

Khosrotash
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Other way is: if we guarantee that the limit exists then: $$\lim_{n\to \infty}\sqrt[n]{n \sqrt[n]{n \sqrt[n]{n ...}}}=L$$ $$\lim_{n\to \infty}\sqrt[n]{n}L^{1/n}=L$$ concludes that $L= 1$

AsdrubalBeltran
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We want to show that $n^{1/(n-1)} \to 1 $.

Since $(1+1/\sqrt{n})^n \ge 1+\sqrt{n} \gt \sqrt{n}$ by Bernoulli's inequality, $n^{1/n} \lt (1+1/\sqrt{n})^2 \lt 1+3/\sqrt{n} $

Since $(1+x)^n \ge 1+nx $, $(1+nx)^{1/n} \le 1+x $ or $(1+x)^{1/n} \lt 1+x/n $.

Therefore

$\begin{array}\\ n^{1/(n-1)} &=n^{1/n-1/n+1/(n-1)}\\ &=n^{1/n}n^{-1/n+1/(n-1)}\\ &=n^{1/n}n^{1/(n(n-1))}\\ &\lt (1+3/\sqrt{n})(1+(n-1))^{1/(n(n-1))}\\ &\le (1+3/\sqrt{n})(1+(n-1)/(n(n-1))\\ &= (1+3/\sqrt{n})(1+1/n) &\to 1\\ \end{array} $

marty cohen
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We have

$$1\le x_n = n^{1/n+1/n^2 +1/n^3 +\cdots + 1/n^n} \le n^{1/n+(n-1)/n^2}\le n^{2/n} = (n^{1/n})^2 \to 1^2 = 1.$$

By the squeeze theorem the limit is $1.$

zhw.
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