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Show lim (as n->inf) of 1/n*{2n/(2n+1)}^2 exists without finding limit

  1. Sep 6, 2014 #1
    1. The problem statement, all variables and given/known data
    Show that lim(n→infinity) 1/n*{2*4*...*2n/(1*3*...*(2n-1))}^2 exists without finding the limit.

    2. Relevant equations
    probably the following:
    Let {xn} be a sequence such that xn≥xn+1 and xn≥M for every n. Then the series is convergent.

    3. The attempt at a solution
    I know I need to do a proof.
    n≥1 (n cannot be 0 because 1/n is undefined and n must be an integer since it's the term number)
    for any integer n>1, ((2n-2)*2n)/(2n-1)2*2n=((2n-1)2-1)/(2k+1)2<1
    Therefore, {xn}2<4n and 1/n*{xn}2<4
    (Hence, I can prove n≥1 and n≤4 (for n=1, 1/1*{2/1)^2=4 so I included 4)
    for n=1 1/1*{2/1}^2=4 and for n=2 1/2*{2*4/(1*3)}^2=64/18=32/9
    thus, n=1>n=2 establishing a basis for an induction proof
    assume for n=k that 1/k*{2*4*...*2k/(1*3*...*(2k-1))}^2>1/(k+1)*{2*4*...*2(k+1)/(1*3*...*(2k))}^2 so that 1/n*{2*4*...*2n/(1*3*...*(2n-1))}^2>1/(n+1)*{2*4*...*2(n+1)/(1*3*...*(2n))}^2

    This is where I am stuck. I want to show that the statement 1/n*{2*4*...*2n/(1*3*...*(2n-1))}^2>1/(n+1)*{2*4*...*2(n+1)/(1*3*...*(2n))}^2 holds true for n=k+1 but I can't figure out how to get to that point. I know you can't just replace the n's with k+1's in the proof.
  2. jcsd
  3. Sep 6, 2014 #2

    Ray Vickson

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    Try using Stirling's formula, together with
    [tex] 1 \cdot 3 \cdot 5 \cdots (2n-1) = \frac{(2n)!}{2 \cdot 4 \cdots 2n} \\
    2 \cdot 4 \cdots 2n = 2^n \times 1 \cdot 2 \cdot 3 \cdots n = 2^n \, n! [/tex]
    This would prove the existence of a limit by actually finding it. However, how you do the problem will depend on what tools and results you are allowed to use.
  4. Sep 6, 2014 #3
    Sometimes it's easier to show ##\frac{a}{b}\geq 1## or ##\frac{b}{a}\leq 1## than it is to show ##a\geq b## (even though these inequalities are equivalent when ##a,b>0##). This is definitely one of those times. You really don't need to use anything fancy like induction or Stirling's formula here. Simple algebra will suffice.

    Also, in order to use the theorem that you're trying to use (the one that says a decreasing sequence that is bounded below must converge), you need to show that the sequence has a lower bound. You gave an argument for why the terms of sequence are only defined when the index is nonzero.

    And your argument (which is a little bit difficult to follow) for the fact that the sequence is bounded above by ##4## is not really necessary given the theorem that you're trying to use.
  5. Sep 6, 2014 #4
    Thanks for the advice. It really helped. I can't believe I didn't think of such a simple solution. I ended using algebra to show xn>xn+1. Then I hashed out my thoughts about xn<4 and used the theorem I was trying to use.
  6. Sep 6, 2014 #5
    It's a standard trick that you want to keep in your back pocket when trying to prove an inequality. Showing ##a-b\geq 0## is another commonly used path towards proving ##a\geq b## that you'll want to keep handy as well.

    Well, like I said, that fact is completely irrelevant to proving what you're trying to prove. Furthermore, given that your sequence is decreasing (which have now proven), ##x_n\leq 4## for all ##n## follows from ##x_1=4##, which is basic arithmetic.
  7. Sep 7, 2014 #6
    Yes, but I wanted to make sure I knew how to prove it if I ever needed to. It seems as I get farther into mathematics I forget about using the basics.
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