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

Question

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

I know how to solve when all surds are of the same order, but what if they are different?

Technically, (as some users wanted to know exactly what is to be found), find:

$$\lim_{n\to\infty}\sqrt{4+\sqrt[3]{4+\sqrt[4]{4+\sqrt[5]{4+\cdots+\sqrt[n]{4}}}}} $$

@Crostul Not really. Although the same argument may apply somewhat. — Riccardo Orlando, May 12 '16 at 07:46
Do you have a reason to expect this number to have a nice closed form? — Wojowu, May 12 '16 at 08:00
@wojowu not sure.. this question is from a math competition paper.. — Nitin Singh, May 12 '16 at 08:06
@Apurv Fiitjee math recruitment test paper. You perhaps know about fiitjee — Nitin Singh, May 12 '16 at 08:14
If this is from a Fiitjee math recruitment test paper, then it most likely has a closed form. — Aritra Das, May 12 '16 at 08:18
Could we maybe construct a recurrence relation ? And then find the limit as $n\to\infty$ from there ? — Aritra Das, May 12 '16 at 08:27
Perhaps this is obvious to everybody else, but how exactly do you define the value of that expression? — Martin R, May 12 '16 at 09:48
@MartinR I think, as the limit (assuming it exists) $$\lim_{n\to\infty}\sqrt{4+\sqrt[3]{4+\sqrt[4]{4+\sqrt[5]{4+...\sqrt[n]{4}}}}} $$ — Aritra Das, May 12 '16 at 09:59
@NitinSingh Do you know the answer to the question? (Not the decimal approximation. The exact closed form) — Aritra Das, May 12 '16 at 10:16
I found this to be thought provoking, although to no avail. Maybe someone abler than me will succeed in putting it to use: $$x^{e-2} = \sqrt{x\sqrt[3]{x\sqrt[4]{x\sqrt[5]{x\dots}}}}$$ — Aritra Das, May 12 '16 at 20:19
The value at least exists since the sequence is monotonically increasing and bounded by $\sqrt{4 + \sqrt{4 + \sqrt{4 + ...}}}$. — Jack M, May 13 '16 at 00:41

orion · Answer 1 · 2016-05-13T12:12:07.083

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?

I think you made a mistake. The sequence should converge for most values of $x$. See Jack M's comment above. — Quang Hoang, May 13 '16 at 07:28
I inverted the sequence - I'm starting with the unknown and computing the innermost terms... if you start in the innermost sqrt and go outwards, it converges for all initial points (but the problem is you can't start at infinity, so you need to express it as a limit we don't know what to do with). If you go from the outside it's completely unstable so there is only one $x$ (value of the expression) that traverses the iteration in reverse. — orion, May 13 '16 at 08:44

score 1 · Answer 2 · answered May 14 '16 at 17:36

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$$y\approx\sqrt{4+\left(4+\left(4+\left(4+\left(4+\sqrt[6]4\right)^{1/6}\right)^{1/5}\right)^{1/4}\right)^{1/3}}\approx \color{red}{2.40161550315}$$

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

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