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An Interesting Limit.

  1. Mar 18, 2012 #1
    I've recently been confronted with the limit as n goes to 0 of sin([itex]\pi[/itex]*n/4)*[itex]\Gamma[/itex](x) , and have no idea on how to confront the problem, as I have little familiarity with the gamma function. Is there any relatively easy ways to prove this, or at least ways that use methods not difficult to learn? I would very much like to see a proof, as wolfram alpha gives a answer of [itex]\pi[/itex]/4, and the answer is important relating to some very interesting alternating series.
    edit 2: Wow, I haven't solved this problem, but if wolfram alpha is right, soon I'll be summing alternating series never summed before :) Well, ones I've never seen summed before at least...
     
    Last edited: Mar 18, 2012
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  3. Mar 22, 2012 #2

    morphism

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    ##\lim_{n \to 0} \sin(\frac{\pi n}{4}) \Gamma(x) = \sin(0)\Gamma(x) = 0##...
     
  4. Mar 23, 2012 #3

    mathman

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    I suspect there is a typo in the original question - it should be Γ(n), not Γ(x).
     
  5. Mar 23, 2012 #4
    It's actually quite easy. You know that the gamma function satisfies the functional equation [itex]\Gamma(x+1) = x \Gamma(x)[/itex], so then:

    [tex]\begin{align*} \lim_{x \rightarrow 0} \sin (\pi x/4) \Gamma(x) &= \lim_{x \rightarrow 0} \frac{\sin(\pi x/4) \Gamma(x+1)}{x} \\ &= \left( lim_{x \rightarrow 0} \frac{\sin(\pi x/4)}{x} \right) \left( \lim_{x \rightarrow 0} \Gamma(x+1) \right) \\ &= lim_{x \rightarrow 0} \frac{\sin(\pi x/4)}{x}\end{align*}[/tex]

    Where the last equality follows from the fact that [itex]\Gamma[/itex] is continuous and [itex]\Gamma(1) = 1[/itex]. But then by L'hopital's rule:

    [tex]\lim_{x \rightarrow 0} \frac{\sin(\pi x/4)}{x} = \lim_{x \rightarrow 0} \frac{\pi}{4} \cos (\pi x/4) = \frac{\pi}{4}[/tex]
     
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