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

  1. Mar 15, 2016 #1
    1. The problem statement, all variables and given/known data
    Let (an)n≥1 be a sequence with a1≥0 and an+1=an(an+4), n≥1. Compute limn→∞ (an)(1/(2n)).

    2. Relevant equations
    a1≥0
    an+1=an(an+4), n≥1
    L = limn→∞ (an)(1/(2n))

    3. The attempt at a solution
    Firstly, I had tried to see if an can be expressed only in terms of a1, but I couldn't get something out of this idea.

    Then, I used the root criterion:
    an+1/an = an+4 → 4 + l, where limn→∞ an = l ∈ℝ∪{-∞,∞}
    ⇒ limn→∞ (an)1/n = 4 + l
    But L= limn→∞ (an)(1/(2n)) = (limn→∞ (an)1/n)limn→∞ 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 limn→∞ an = +∞ (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= limn→∞ (an)(1/(2n)) = limn→∞ eln[(an)(1/(2n))] = elimn→∞ (ln(an))/(2n)
    So everything gets down to computing L'=limn→∞ (ln(an))/(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?
     
  2. jcsd
  3. Mar 15, 2016 #2

    mfb

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    2016 Award

    Staff: Mentor

    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.
     
  4. Mar 15, 2016 #3

    pasmith

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    Homework Helper

    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].
     
    Last edited: Mar 15, 2016
  5. Mar 15, 2016 #4

    mfb

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    2016 Award

    Staff: Mentor

    If we get that, where is the point in bn? 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.
     
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