Find $\sqrt{4+\sqrt[3]{4+\sqrt[4]{4+\sqrt[5]{4+\cdots}}}}$

Solution 1:

It's a bit longer comment so I'll write here (but it's not a solution so I'm deleting when a real answer appears).

I'm thinking along the lines of the sequence $$f_1=x+4$$ $$f_2=(f_1-4)^2$$ $$f_n=(f_{n-1}-4)^n$$ This is the sequence of radicands, which is divergent for every $x$ except for the one that solves our question. So we're looking for $x$ for which this sequence is monotonically convergent down to $5$. Any minor deviation, and it starts diverging. So we're looking for an unstable critical point of the sequence above.

$(f_n)$ is a sequence of polynomials in $x$: $$x+4,x^2,x^6-12x^4+48x^2-64,x^{24}-48x^{22}+\cdots$$

The solution is $x$, such as $\lim_{n\to\infty}f_n=5$ exists, but we could restate it as a convergence criterion $\lim_{n\to\infty}\frac{f_{n+1}}{f_n}= 1$. I was hoping that in the limit, only a subset of polynomial coefficients with known asymptotics matter, but here I got stuck, because it appears all of them are important.

Another way of looking at this is to find the largest root of $f_n-5=0$ in the limit $n\to\infty$.

I figured out the first order asymptotics for $f_n$ at the correct $x$:

$$f_n\asymp 5 + \frac{\ln 5}{n}$$

Any thoughts on this approach?

Solution 2:

Put $$y=\sqrt{4+\sqrt[3]{4+\sqrt[4]{4+\sqrt[5]{4+...}}}}\qquad (1)$$ stopping successively at the $n$-th root we have a sequence strictly increasing and bounded so the limit $y$ is well defined.

From $(1)$ we can easily get a sequence $\{P_n\}$ of polynomials such that its largest real roots $$\alpha_n=\sqrt{4+(4+(4+(4+….(4+\sqrt[n]4)^{1/n}….)^{1/5})^{1/4})^{1/3}}$$ form a sequence $\{\alpha_n\}$ converging to $y$. In fact we have $$\begin{cases}P_2(x)=x^2-4\\P_3(x)=(x^2-4)^3-4\\P_4(x)=( (x^2-4)^3-4)^4-4\\.....\\.....\\P_n(x)=(P_{n-1}(x))^n-4\end{cases}$$

We can see the polynomial $P_n$ has degree $n!$ and $P_n(\alpha_{n-1})=-4$. Since $P_n(x)$ grows very quickly, this indicates that $\{\alpha_n\}$ rapidly converges to the limit $y$ because the arc of the graphic of $P_n(x)$ between the points $(\alpha_{n-1},-4)$ and $(\alpha_n,0)$ is almost a vertical line so $\alpha_{n-1}\approx \alpha_n$. This argument is enhanced by the equality $$y=2\sqrt{1+\left(\frac{4}{4^3}+\left(\frac{4}{4^{12}}+\left(\frac{4}{4^{60}}+….\left(\frac{4}{4^n}+\sqrt[n]4\right)^{1/n}….\right)^{1/5}\right)^{1/4}\right)^{1/3}}$$ hence the factor of $2$ is equal to $$\sqrt{1+\left(\frac{1}{2^{6-2}}+\left(\frac{1}{2^{24-2}}+\left(\frac{1}{2^{120-2}}+….\left(\frac{1}{2^{n!-2}}+\sqrt[n]4\right)^{1/n}….\right)^{1/5}\right)^{1/4}\right)^{1/3}}$$ or $$\sqrt{1+\left(\frac{1}{2^{4}}+\left(\frac{1}{2^{22}}+\left(\frac{1}{2^{118}}+….\left(\frac{1}{2^{n!-2}}+\sqrt[n]4\right)^{1/n}….\right)^{1/5}\right)^{1/4}\right)^{1/3}}$$ This factor of $2$ grows very slowly obviously.

It seems to me there is not a closed form for $y$ (It could be an asymptotic one perhaps). Consequently I give here an (justified) approximation taking the above referred real root of, say, $P_6$; we have$$y\approx\sqrt{4+\left(4+\left(4+\left(4+\left(4+\sqrt[6]4\right)^{1/6}\right)^{1/5}\right)^{1/4}\right)^{1/3}}\approx \color{red}{2.40161550315}$$