# The 'Factorialth Root'

I was dealing with the following question, given by my friend:

Let $\xi(x)=\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{\cdots}}}}$

Define the series $X$ as $\xi(1),\xi(2),\xi(3),\dots$

Find $n$ for which $\xi(n)$ is the 51st Whole Number in the series.

I solved it, of course, [and interestingly $\xi(1)={{1+\sqrt5}\over2}$, the Golden Ratio] but that led us on a competition in which we would try to find out the value of increasingly convoluted expressions.

Some time later, I made an expression, which I called 'The Factorialth Root', written as $\sqrt[!]{x}$.

For some $x$, $\sqrt[!]{x}=\sqrt{x\sqrt{(x-1)\sqrt{(x-2)\sqrt{\ddots\sqrt{2\sqrt1}}}}}$

My friend thought that $(x>y)\to(\sqrt[!]{x}<\sqrt[!]{y})$, while I thought the opposite, that $(x>y)\to(\sqrt[!]{x}>\sqrt[!]{y})$.

I showed by example that mine was correct, but couldn't prove it.

My attempt:

If $[(x>y)\to(\sqrt[!]{x}>\sqrt[!]{y})]$ is true, then $\sqrt[!]{x}>\sqrt[!]{x-1}$. This is possible only when $x>\sqrt[!]{x-1}$. It follows that $\sqrt[!]{2}>\sqrt[!]{1},\sqrt[!]{3}>\sqrt[!]{2}$, and so on.

So, I thought I could prove it by induction, but can't seem to find any way to apply it.

Can anyone help?

For $n > 1$,

\begin{align} \frac{\sqrt[!]{n}}{\sqrt[!]{n-1}} &= \frac{n^{1/2} \cdot (n-1)^{1/4} \cdot (n-2)^{1/8} \cdot\,\cdots\,\cdot 1^{1/2^{n\phantom{-1}}}}{\phantom{n^{1/2}\cdot}(n-1)^{1/2}\cdot(n-2)^{1/4}\cdot\,\cdots\,\cdot 1^{1/2^{n-1}}} \\[4pt] &= \frac{n^{1/2}}{(n-1)^{1/4}\cdot(n-2)^{1/8}\cdot\,\cdots\,\cdot 1^{1/2^n}} \\[4pt] &= \frac{n^{1/4+1/8+1/16+\cdots+1/2^{n}+1/2^{n}}}{(n-1)^{1/4}\cdot(n-2)^{1/8}\cdot\,\cdots\,\cdot 1^{1/2^n}} \\[4pt] &= \left(\frac{n}{n-1}\right)^{1/4}\left(\frac{n}{n-2}\right)^{1/8}\left(\frac{n}{n-3}\right)^{1/16}\cdot\,\cdots\,\cdot \left(\frac{n}{1}\right)^{1/2^{n}}\cdot n^{1/2^n} \\[4pt] &> 1 \cdot 1 \cdot 1 \cdot\,\cdots\,\cdot1 \cdot 1 \\[4pt] &= 1 \end{align}

Let's define $a_n=\sqrt[!]{n}$ for easier notation.

You have

$$a_n=\sqrt{n a_{n-1}}$$ in other words, the next term is the geometric mean of $n$ and previous term. This implies the order (because the mean is always between the values):

$$n>a_n>a_{n-1}\quad \text{if}\quad n>a_{n-1}$$ or $$n<a_{n}<a_{n-1}\quad \text{if}\quad n<a_{n-1}$$

Which one is it? It's easy to show that if the condition $n>a_{n-1}$ holds for one term, it will hold for all next terms also (this is almost trivial induction, because the ordering will say $n>a_n$ which of course implies $n+1>a_n$). You have $2>a_1=1$, so this proves it.

You can explore this further. You may notice that after $n=4$, you get $n-2<a_n<n-1$. In fact, you can show that $a_n$ comes arbitrarily close to $n-1$. Conjecture: $$\lim_{n\to \infty}\left[(n-1)-a_n\right]=0$$ Define the sequence of differences:

$$b_n=(n-1)-a_n$$ Next step: $$b_{n+1}=n-\sqrt{(n+1) a_n}$$ $$b_{n+1}=n-\sqrt{(n+1)(n-1-b_n)}=n-\sqrt{n^2-1-b_n(n+1)}=n(1-\sqrt{1-\frac{b_n}{n}-\frac{b_n+1}{n^2}})$$ Now first use the inequality $\sqrt{1+x}<1+\frac{x}{2}$ which you can easily prove by squaring. $$b_{n+1}<n(1-1+\frac{b_n}{2n}+\frac{b_n+1}{2n^2})=\frac{b_n}{2}+\frac{b_n+1}{2n}=b_{n}\left(\frac{1}{2}+\frac{1}{n}\right)+\frac{1}{2n}$$ For large $n$, the second term gets arbitrarily small and you can show that $b_{n}$ sequence decreases faster toward zero than any geometric series with factor $q>\frac{1}{2}$ after some suitably large $n$. Or, to put it differently, the ratio test shows convergence of $b_{n+1}$ toward zero:

$$\frac{b_{n+1}}{b_n}=\frac{1}{2}+\frac{1}{n}+\frac{1}{2nb_{n}}$$ Once you find one $b_n$ for which the right side of equation is $<1$, it holds forevermore. And you get there at $n=4$.

This means that for large $n$, you have asymptotic behaviour $a_n\asymp (n-1)$.