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Series Calc II

  1. May 1, 2009 #1
    Find the largest value of b that makes the following statement true: "if 0<= a <= b, then the series (from n=1 to infinity) of (((n!)^2a^n)/(2n!)) converges".

    I know you have to do the ratio test for this one but I don't know how to do it.
     
  2. jcsd
  3. May 1, 2009 #2

    Tom Mattson

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    OK let's call the summand [itex]s_n[/itex]. So we have:

    [tex]s_n=\frac{n!^{{2a}^n}}{(2n)!}[/tex]

    Can you write down [itex]s_{n+1}[/itex]?
     
  4. May 1, 2009 #3
    im sry its (n!)^2(a^n) for the numerator
     
  5. May 1, 2009 #4

    HallsofIvy

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    Yes, use the ratio test.
    [tex]\left[\frac{((n+1)!)^2a^{n+1}}{(2(n+1))!}\right]\left[\frac{(2n)!}{((n!)^2a^n}\right]= \left[\frac{(n+1)!}{n!}\right]^2\left[\frac{a^{n+1}}{a^n}\right]\left[\frac{(2n)!}{(2(n+1))!}][/tex]
    [tex]= \frac{(n+1)(a)}{(2n+2)(2n+1)[/tex]
    What is the limit of that as n goes to infinity? If that limit depends on a, then the series will converge only for values of a that make the limit less than 1.
     
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