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Show Convergence of sequence (1 + c)(1 + c^2) (1 + c^n)

  1. Jan 24, 2012 #1
    1. The problem statement, all variables and given/known data

    Hey, so here is the problem:

    Suppose 0≤c<1 and let an = (1 + c)(1 + c2)...(1 + cn) for integer n≥1. Show that this sequence is convergent.

    Well I understand the basic concepts of proving convergence of sequences, but in class we've only ever done it with sequences where the terms are summed not multiplied like this one is. I guess I'm just having trouble figuring out where to go really...

    2. Relevant equations

    I guessed we either need to use the theorem that says any bounded monotone sequence is convergent.

    Or by simply finding the limit L and proving that the sequence converges to L.

    3. The attempt at a solution

    There's a suggestion that says " Show that 1 + c ≤ ec ".

    I did that by showing that if c = 0, then 1 + 0 = e0 = 1

    And then for any c>0 d/dc(1 + c) < d/dc(ec).

    Therefore for any c≥0, 1 + c ≤ ec.

    Now I don't really know how to use this to solve my problem.

    I tried to show that the sequence is monotone:

    As an+1 = an(1+cn+1)

    Therefore an+1 ≥ an

    Therefore an is non-decreasing, therefore monotone.

    But now I have no idea what to do. I really need help cause my prof just says that it should be obvious.
     
    Last edited: Jan 24, 2012
  2. jcsd
  3. Jan 24, 2012 #2

    tiny-tim

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    welcome to pf!

    hi looserlama! welcome to pf! :smile:
    hint: 1 + c2 ≤ … ? :wink:
     
  4. Jan 24, 2012 #3
    thanks tiny-tim.

    Would it not be 1 + c2 ≤ ec2 ?

    Does that just mean (1 + c)(1 + c2)...(1 + cn) ≤ ecec2...ecn = ec + c2 + ... + cn = e(c - cn+1)/(1 - c)

    That's probably taking it too far, I just don't really understand where to go with that?
     
  5. Jan 24, 2012 #4

    Dick

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    That's not taking it too far. Take it even farther. e^((c - c^(n+1))/(1 - c)))<=e^(c/(1-c)). So your products are bounded from above. Now what?
     
  6. Jan 24, 2012 #5
    Oh yea, so if an is smaller than a finite number, then it's bounded from above and given an ≥ 0, it's bounded. Therefore it's convergent.

    But I need to prove that 1 + cn ≤ ecn for any integer n ≥ 1 .

    I don't know how to do that?

    The only way I can think of is with induction, so this is what I have:

    Basic Step: For n = 0

    1 + c ≤ ec

    Inductive Step:

    If the statement holds for integer k ≥ 1, then

    1 + ck ≤ eck

    so 1 + ck + 1 ≤ ceck

    but how do I show that ceck ≤ eck + 1 ?
     
    Last edited: Jan 24, 2012
  7. Jan 24, 2012 #6

    Dick

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    Now you are deviating from the correct path. Sure, an<=e^(nc). But that's no use. As n->infinity e^(nc) goes to infinity. Bad. And proving 1+c^n<e^(c^n) is no real trick if you've already proved 1+c<e^c. Just substitute c^n for c!
     
  8. Jan 24, 2012 #7
    Yea I saw that what I wrote before was wrong, so I edited my previous post.

    I just am having trouble rigorously proving that 1 + c^2 <= e^(c^n). See my edited post for what I did.
     
  9. Jan 24, 2012 #8

    Dick

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    If 1+c<=e^(c) then 1+c^2<=e^(c^2). Just substitute c^2 for c in the first equation. I told you that. You don't need a separate proof if you already proved the first one.
     
  10. Jan 25, 2012 #9
    Oh I see. U was thinking about that the wrong way.

    Thanks Dick
     
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