A limit involving a recurrent sequence: a(n+1)=a(n)*(a(n)+4)

[anachin6000]

Homework Statement

Let (a_n)_n≥1 be a sequence with a₁≥0 and a_n+1=a_n(a_n+4), n≥1. Compute lim_n→∞ (a_n)^(1/(2n)).

Homework Equations

a₁≥0
a_n+1=a_n(a_n+4), n≥1
L = lim_n→∞ (a_n)^(1/(2n))

The Attempt at a Solution

Firstly, I had tried to see if a_n can be expressed only in terms of a₁, but I couldn't get something out of this idea.

Then, I used the root criterion:
a_n+1/a_n = a_n+4 → 4 + l, where lim_n→∞ a_n = l ∈ℝ∪{-∞,∞}
⇒ lim_n→∞ (a_n)^1/n = 4 + l
But L= lim_n→∞ (a_n)^(1/(2n)) = (lim_n→∞ (a_n)^1/n)^{lim_n→∞ n/(2n)} = (4 + l)⁰
So:
- if l∈ℝ\{-4}, then L=1
- and if l∈{-∞, -4,∞}, then L is a ∞⁰ or 0⁰ indeterminate.

Though, the sequence is either constant (equal to 0) if a₁ = 0 (so L=1) or has lim_n→∞ a_n = +∞ (a₁ > 0) if, so the job is not done.

Knowing that I have to deal with a ∞⁰, I rewrote the limit as follows:
L= lim_n→∞ (a_n)^(1/(2n)) = lim_n→∞ e^{ln[(a_n)(1/(2n))]} = e^{lim_n→∞ (ln(a_n))/(2n)}
So everything gets down to computing L'=lim_n→∞ (ln(a_n))/(2ⁿ)

This is where I got stuck. I tried to use the root criterion again, I tried Stoltz-Cesaro Theorem, both with no success.

Can anyone help me compute this limit or at least give me a direction to continue?

 

[mfb]

A very interesting problem. With some examples, I found very good approximations for very small and very large a₁, and a good approximation over the whole range, but nothing that would fit exactly.

For very large a₁, it is possible to identify the leading contributions (the highest powers of a), and develop a formula for the limit. That is not exact, but the error goes to zero for a₁ to infinity.

 

[pasmith]

It helps to write [itex]a_{n+1} = a_n^2(1 + \frac{4}{a_n})[/itex] and set [itex]b_n = \frac{\log a_n}{2^n}[/itex] so that [tex]
b_{n+1} - b_n = \frac{1}{2^{n+1}} \log \left( 1 + \frac{4}{a_n} \right).
[/tex] Now [itex]a_n \to \infty[/itex] for any [itex]a_1 > 0[/itex], so there exists [itex]N \in \mathbb{N}[/itex] such that if [itex]n \geq N[/itex] then [itex]a_n > \frac{4}{e^2 - 1}[/itex]. Hence for [itex]n \geq N[/itex] we have [tex]
b_{n+1} - b_n = \frac{1}{2^{n+1}} \log \left( 1 + \frac{4}{a_n} \right) < \frac{1}{2^n}.[/tex] But then [itex]b_n[/itex] is cauchy, and so converges to some finite limit. It is possible that the exact limit depends on [itex]a_1[/itex] though, which is awkward.

EDIT: Since we know [itex]b_{n+1} - b_n[/itex] in terms of [itex]a_n[/itex] we can immediately sum from [itex]n=1[/itex] to [itex]n=N-1[/itex] to obtain [tex]
b_N = b_1 + \sum_{n=1}^{N-1} \frac{1}{2^{n+1}} \log \left( 1 + \frac{4}{a_n} \right)[/tex] and hence write down an expression for [itex]\lim_{n \to \infty} a_n^{1/2^n}[/itex]. Then all we need is an expression for [itex]a_n[/itex] in terms of [itex]a_1[/itex].

 

[mfb]

Then all we need is an expression for [itex]a_n[/itex] in terms of [itex]a_1[/itex].

If we get that, where is the point in b_n? You would replace a direct limit (take that expression to the power of 1/(2n)) with a series.

The limit does depend on a₀ in a nontrivial way.