An asymptotic formula for a nested radical

Question

Consider a nested radical (written with exponents) of the form:

$$F(n,k,q)=\left(k^n+\left(k^{n+1}+\left(k^{n+2}+\left(k^{n+3}+\cdots\right)^{1/q}\right)^{1/q}\right)^{1/q}\right)^{1/q}$$

I have observed that as $n\to\infty$, we have $$F(n,k,q)\sim k^{\frac{n}{q}}+\frac1q k^{\frac{1+n(2-q)}{q}}$$

How can we prove that this asymptotic is true?

Answer 1

First, notice the recursion $\;F(n,k,q) = (k^n\!+\!F(n\!+\!1,k,q))^{1/q}\;$ which we use in the following.

For simplicity let $z := k^{1/q},\; y := z/k,\; x := y^nz.\;$ Define, by recursion, the sequence $\;a_i(t)\;$ with $\;a_0(t) := 1 + O(t),\;$ and $\;a_{i+1}(t) := (1 + q\;t\; a_i(ty))^{1/q}.\;$ Thus, we get $\;a_1(t) = 1 + t + O(t^2),\;$ $\;a_2(t) = 1 + t + (1-q+2y)t^2/2! + O(t^3),\;$ and so on. Now, the asymptotic formula is, for all $i$, $F(n,k,q) \sim z^n a_i(x/q).\;$ For $i=1,\;$ it is $k^{n/q}(1 + x/q + O(x^2)) = k^{\frac{n}{q}}(1 + \frac1q k^{\frac{n+1}{q}-n}+O(x^2)).$

Your nested radical is also called a continued radical and they have a long history.

Edit: An example computation using PARI/GP looks like this:

? nxt(at) = (1 + q*t*subst(at,t,t*Y))^(1/q);
? a0 = 1 + O(t); a1 = nxt(a0); a2 = nxt(a1);
? print("a0 = ",a0); print("a1 = ",a1); print("a2 = ",a2);
a0 = 1 + O(t)
a1 = 1 + t + O(t^2)
a2 = 1 + t + (-1/2*q + (Y + 1/2))*t^2 + O(t^3)