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Is there a general identity for the infinite radicals; $$f(n)=\sqrt{n^{0}+\sqrt{n^{1}+\sqrt{n^{2}+\sqrt{n^{3}+...}}}}$$

For $n=0,1,4$ we get $f(n)=1,φ,2$ respectively (If you allow that $0^0=1)$ but will there be any other value of $n$ such that $f(n)∈ \mathbb{Z}$ or even has a closed-form expression? This is actually such an interesting nested radical, that I was unable to even find a good proof for $f(4)=2$. Most of them were reverse-engineering the radical and showing but not proving.

So is there a closed form expression for this as general or even for any other value of $m$ I missed out?

A C++ code for computing this can be given as;

#include <iostream>
#include <math.h>

using namespace std;

long double seq( unsigned long, long double, unsigned long = 0 );

int main() {
    unsigned long m = 0; long double n;
    cout << "Enter integer r: ";
    cin >> m; cout << "Enter n: ";
    cin >> n; cout << seq( m, n); return 0;
}

long double seq( unsigned long m, long double n, unsigned long i) {
    return sqrt( pow(n,i)+(i==m?0:seq(m,n,i+1)) );
}

Where input $r$ means how many times you need to continue the radicals, and $n$ will be the number you want to put in.

Edit;

Maybe a trivial finding, but I also got that

$$f(n)=\sqrt{1+n^{\delta(1)}\sqrt{1+n^{\delta(2)}\sqrt{1+n^{\delta(3)}\sqrt{1+...}}}}$$

Where,

$$\delta(x)=(1-x)+\frac{1}{2x}$$

user712576
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    Note that computing a sequence with a computer, especially with finite precision and naive implementations is not a proof of convergence. for two reasons: 1- If you are computing $S_n= S_{n-1}+a_n$, once $n$ is big enough such that $a_n< \varepsilon(S_n)$, then the comparison $S_n== S_{n-1}$ yields true even for non zero $a_n$. Also, even if you didn't have precision issues, divergence could be very slow, a famous case is the harmonic series, for which the sum of the first $10^{43}$ terms is less than 100 – Mefitico Oct 24 '19 at 12:26

1 Answers1

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You are studying the limit of nested radicals of the form

$$a_{k} = \sqrt {x^0 + \sqrt{x^1+\cdots +\sqrt{x^k}}}$$

There is a theorem by Herschfeld (On infinite radicals, 1935) that shows $a_k$ converges if and only if the sequence

$$F_k := (x^k)^{1/2^k} = x^{k/2^k}$$

is bounded. This happens for every $x\in \mathbb{N}$.

In the article by Herschfeld there are several results on nested radicals (including the case for $x=n^2$ which comprises your request $x=4$) of this form but also of the "left-hand" form

$$u_k = \sqrt{a_k + \sqrt{a_{k-1}+\ldots +\sqrt{a_0}}}$$

of which the author is able to calculate the general limit in terms of the limit of sequence $\{a_k\}$.

Caligula
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  • Yes, and I also looked into further cases of Herschfeld's Theorem, and I found out that Vijayaraghavan(https://en.wikipedia.org/wiki/Tirukkannapuram_Vijayaraghavan) has proved that a sequence $\ \sqrt{a_1 + \sqrt{a_2 +:\cdots: +\sqrt{a_n}}}\ \ $ converges if and only if $\displaystyle\ \ {\overline {\lim_{n\to\infty}}}\ \frac{\log:{a_n}}{2^n}: < :\infty:.: $ I believe all natural numbers converges, or am i wrong on this? – user712576 Oct 24 '19 at 13:56