Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Proof of Convergent Sequence

  1. Feb 11, 2010 #1
    1. The problem statement, all variables and given/known data

    If [tex]a_{1}[/tex] = 1 and [tex]a_{n+1}[/tex] = (1-(1/[tex]2^{n}[/tex])) [tex]a_{n}[/tex], prove that [tex]a_{n}[/tex] converges.

    2. Relevant equations

    3. The attempt at a solution
    I am confident about my attempt, I just want it checked. Thanks.

    First show that [tex]a_{n}[/tex] is monotone: [tex]a_{n}[/tex] = {1, 1/4, 21/32, 315/512,...}

    Claim: [tex]a_{n+1}[/tex] [tex]\leq[/tex] [tex]a_{n}[/tex] for all n [tex]\in[/tex] N. (Must show this)


    [tex]a_{n+1}[/tex] [tex]\leq[/tex] [tex]a_{n}[/tex]

    [tex]\Leftrightarrow[/tex] (1-(1/[tex]2^{n}[/tex])) [tex]a_{n}[/tex] [tex]\leq[/tex] [tex]a_{n}[/tex]

    [tex]\Leftrightarrow[/tex] [tex]a_{n}[/tex] - ([tex]a_{n}[/tex] / [tex]2^{n}[/tex]) [tex]\leq[/tex] [tex]a_{n}[/tex]

    [tex]\Leftrightarrow[/tex] [tex]2^{n}[/tex] [tex]a_{n}[/tex] - [tex]a_{n}[/tex] [tex]\leq[/tex] [tex]2^{n}[/tex] [tex]a_{n}[/tex]

    [tex]\Leftrightarrow[/tex] [tex]2^{n}[/tex] [tex]a_{n}[/tex] - [tex]2^{n}[/tex] [tex]a_{n}[/tex] [tex]\leq[/tex] [tex]a_{n}[/tex]

    [tex]\Leftrightarrow[/tex] 0 [tex]\leq[/tex] [tex]a_{n}[/tex]

    So we can conclude that [tex]a_{n+1}[/tex] [tex]\leq[/tex] [tex]a_{n}[/tex] as long as [tex]a_{n}[/tex] [tex]\geq[/tex] 0. Now [tex]a_{1}[/tex] = 1, so the sequence starts with a number greater than or equal to zero. Every other number in the sequence has the form:

    x(1-(1/[tex]2^{x}[/tex])) for x greater than zero.

    We claim every such number is greater than or equal to zero, and will show this by induction.

    Induction Step: Assume [tex]a_{n}[/tex] [tex]\geq[/tex] 0

    [tex]a_{n+1}[/tex] = [tex]a_{n}[/tex] (1-(1/[tex]2^{n}[/tex])) [tex]\geq[/tex] 0(1-(1/2)) = 0

    This implies [tex]a_{n+1}[/tex] [tex]\geq[/tex] [tex]a_{n}[/tex], so since [tex]a_{n}[/tex] [tex]\geq[/tex] 0 for all n [tex]\in[/tex] N, this implies that [tex]a_{n}[/tex] is bounded and monotonic. So by the Monotone Sequence Theorem, [tex]a_{n}[/tex] converges for all n [tex]\in[/tex] N ■
  2. jcsd
  3. Feb 11, 2010 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Your proof looks reasonable.

    You misstated the conclusion though -- For each n, an is a number so it doesn't make sense to say "an converges for all n". What you meant is that the sequence an converges.
  4. Feb 11, 2010 #3
    ah yes, thanks for the tip.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook