Why does $\sqrt{n\sqrt{n\sqrt{n \ldots}}} = n$?

Question

Ok, so I've been playing around with radical graphs and such lately, and I discovered that if the

nth x = √(1st x √ 2nd x ... √nth x);

Then

$$\text{the "infinith" } x = x$$

Example:

$$\sqrt{4\sqrt{4\sqrt{4\sqrt{4\ldots}}}}=4$$
Try it yourself, type calc in Google search, hit then a number, such as $4$, and repeat, ending with $4$, (or press the buttons instead).

I'm a math-head, not big enough though, I think this sequence is divergent or convergent or whatever, too lazy to search up the difference.

However, can this be explained to me? Like how the Pythagorean Theorem can be explained visually.

Answer 1

Suppose $y = \sqrt{x\sqrt{x\sqrt{x\cdots}}}$.

Multiply both sides by $x$ and take the square root:

$$\sqrt{xy} = \sqrt{x\sqrt{x\sqrt{x\cdots}}} = y$$

Therefore, $\sqrt{xy} = y$, and solving we have $xy = y^2 \implies x = y$.

Answer 2

You're basically doing this:

$x_0 = \sqrt x$

$x_1 = \sqrt{x x_0}$

$\displaystyle x_n = \sqrt{x x_{n-1}} = x^{\sum_{k=1}^n \frac{1}{2^k}} $

So it's pretty obvious that converges to $x$

Answer 3

$$\begin{align} X & = \sqrt{n \cdot \sqrt {n \cdot \sqrt{n \dots} } } \\ & = \sqrt{n} \cdot \sqrt{\sqrt{n}} \cdot \sqrt{\sqrt{\sqrt{n}}} \cdot \dots \\ &= n^{1/2} \cdot n^{1/4} \cdot n^{1/8} \dots \\ &= n^{1/2 + 1/4 + 1/8 \dots} \\ &= n^1 \\ &= n \\ \end{align}$$

Answer 4

Suppose $y = \sqrt{x\sqrt{x\sqrt{x\cdots}}}$. Then obviously $y^2=xy$, whence $y=x$ or $y=0$. But $y \gt 0$, so $y=x$.

Answer 5

It is important to show that the limit exists. Let define the sequence $$ a_k=\sqrt{\vphantom{A}na_{k-1}} $$ Since $\dfrac{a_k}{a_{k-1}}=\sqrt{\dfrac{n}{a_{k-1}}}$ and $\dfrac{a_k}{n}=\sqrt{\dfrac{a_{k-1}}{n}}$, we have

1. if $a_{k-1}\le n$, then $a_{k-1}\le a_k\le n$; that is, $a_k$ is increasing and bounded above by $n$.

2. if $a_{k-1}\ge n$, then $a_{k-1}\ge a_k\ge n$; that is, $a_k$ is decreasing and bounded below by $n$.

In either case, $a_k$ is convergent. Using the continuity of multiplication by a constant and the continuity of square root, we get $$ \lim_{k\to\infty}a_k=\lim_{k\to\infty}\sqrt{\vphantom{A}na_{k-1}}=\sqrt{n\lim_{k\to\infty}a_k} $$ Squaring and dividing by $\lim_{k\to\infty}a_k$, we get that $$ \lim_{k\to\infty}a_k=n $$

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Another Approach

Not as rigorous, but perhaps more intuitive. Take the logarithm of both sides and we get $$ \begin{align} \log\left(\sqrt{n\sqrt{n\sqrt{n\dots}}}\right) &=\frac12\left(\log(n)+\frac12\left(\log(n)+\frac12\left(\log(n)+\vphantom{\frac12}\dots\right)\right)\right)\\ &=\frac12\log(n)+\frac14\log(n)+\frac18\log(n)+\dots\\ &=\log(n)\left(\frac12+\frac14+\frac18+\dots\right)\\[6pt] &=\log(n) \end{align} $$

Answer 6

Assume that you iterated infinitely many times (can take a while), and observed a convergence to $n$.

One more iteration yields $\sqrt{n.n}=n$.