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Homework Help: Squeeze theorem and sequences

  1. Oct 19, 2012 #1
    1. The problem statement, all variables and given/known data

    I have to proof that the sequence (2^n +n^2)/(3^n + 5n^4) converges en calculate its limit using the sqeeuze theorem.

    2. Relevant equations
    (2^n +n^2)/(3^n + 5n^4)


    Theorem 1: Let p[itex]\in[/itex]2N en x[itex]\in[/itex]R with |x|< 1. Then the limit as n goes to infinity of (n^p)(x^n)=0

    3. The attempt at a solution

    I have noticed that if I divide the top and bottom by 3^n then i can use theorem 1 to calculate the limits of the top and the bottom.. the limit at the top will go to 0 and the bottom 1 .. so the limit of the whole goes to 0. My problem is that I dont know in what way i could use the squeeze theorem to proof this.
    Last edited: Oct 19, 2012
  2. jcsd
  3. Oct 19, 2012 #2


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    To use the squeeze theorem, try breaking the fraction into two parts:
    [tex]\frac{2^n + n^2}{3^n + 5n^4} = \frac{2^n}{3^n + 5n^4} + \frac{n^2}{3^n + 5n^4}[/tex]
    Now try bounding each fraction on the right hand side by a simpler expression whose limit is easy to find.
  4. Oct 19, 2012 #3
    Something like this actually came into my head when i was pondering upon the question on the balcony.. : d If this is how long things will take on the exam. I am doomed. Thanks for the tip. I think i will manage to work this out.
  5. Oct 19, 2012 #4
    Actually, I cant see any way to solve this without using theorem 1. I cant find an upper bound for the two fractions.
  6. Oct 19, 2012 #5


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    I'll show you how to find an upper bound for the first fraction. You can do something very similar for the second one. Notice that [itex]5n^4 > 0[/itex], so [itex]3^n + 5n^4 > 3^n[/itex]. Therefore,

    [tex]\frac{2^n}{3^n + 5n^4} < \frac{2^n}{3^n} = \left( \frac{2}{3}\right)^n [/tex]
    Also noting that the fraction is always positive, we have a lower bound of 0, so we may write
    [tex]0 < \frac{2^n}{3^n + 5n^4} < \left( \frac{2}{3}\right)^n[/tex]
    The right-hand side goes to zero as [itex]n \rightarrow \infty[/itex], so the squeeze theorem applies.
  7. Oct 19, 2012 #6
    Thank you a lot, I see it now : )
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