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Convergence/divergence of series

  1. Aug 21, 2005 #1
    I'm having trouble determine whether the following series converge or diverge. I'm supposed to use the ratio test for these.

    [tex]
    \sum\limits_{n = 1}^\infty {\frac{{n!}}{{n^n }}}
    [/tex]

    [tex]
    \sum\limits_{n = 1}^\infty {\frac{{n^2 }}{{e^n }}}
    [/tex]

    For the first one I tried:

    [tex]
    \left| {\frac{{a_{n + 1} }}{{a_n }}} \right| = \left| {\frac{{\left( {n + 1} \right)!}}{{n^{n + 1} }}\frac{{n^n }}{{n!}}} \right| = \frac{{n + 1}}{n} \to 1
    [/tex] as n tends to infinity so wouldn't the ratio test be inconclusive because the limit is 1? The answer tells me that the limit is supposed to be exp(-1) which is less than one so the series converges. The answer for the second one is the same. So I try to manipulate the summations so that they look like the one for the exponential.

    [tex]
    e^x = \sum\limits_{n = 0}^\infty {\frac{{x^n }}{{n!}}}
    [/tex]

    [tex]
    = \sum\limits_{n = 0}^\infty {\left( {\frac{{n!}}{{x^n }}} \right)^{ - 1} }
    [/tex]

    [tex]
    e^n = \sum\limits_{n = 0}^\infty {\left( {\frac{{n!}}{{n^n }}} \right)^{ - 1} }
    [/tex]

    But how would I find the limit of (a_n+1)/(a_n) to be exp(-1)? Can someone help me out?
     
  2. jcsd
  3. Aug 21, 2005 #2

    Hurkyl

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    Staff Emeritus
    Science Advisor
    Gold Member

    You try checking your work for silly mistakes? It's one of the first things I do when I think that the answer I'm getting is wrong.
     
  4. Aug 21, 2005 #3

    LeonhardEuler

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    Gold Member

    Take another look at your ratio test work. You also made a mistake in the last line of your post, but it won't matter because you won't need those equations. Your overworking the variable n.
     
  5. Aug 21, 2005 #4
    Ok I see where I went wrong, thanks for the help.

    [tex]
    \left| {\frac{{a_{n + 1} }}{{a_n }}} \right| = \left| {\frac{{\left( {n + 1} \right)!}}{{\left( {n + 1} \right)^{n + 1} }}\frac{{n^n }}{{n!}}} \right| = \frac{{\left( {n + 1} \right)n^n }}{{\left( {n + 1} \right)\left( {n + 1} \right)^n }} = \left( {\frac{n}{{n + 1}}} \right)^n
    [/tex]

    [tex]
    = \left( {1 + \frac{{ - 1}}{{n + 1}}} \right)^n = \left[ {\left( {1 + \frac{{ - 1}}{{n + 1}}} \right)^{n + 1} } \right]^{\frac{n}{{n + 1}}} \to e^{ - 1}
    [/tex] as n tends to infinity.

    I checked my working numerous times but for some reason I just kept on missing the n^n error when I replaced n by (n+1). By the way, what was my mistake in the last line of my previous post? I think it would be helpful to me if I know what it is. After all, I might need to use it again some day.

    Edit: Nevermind my last sentence. I know what you mean now LE. :blushing:

    Edit 2: Does the product of limits apply when a limit to infinity is taken? For finite a and continuous functions f and g, lim(x->a)(fg) = (lim(x->a)f)(lim(x->a)g) if I recall correctly. Does this apply if a approaches infinity?
     
    Last edited: Aug 21, 2005
  6. Aug 21, 2005 #5

    lurflurf

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    Homework Helper

    Yes, when all three limits exist. Limits don't care about the value of a only what the function does. You can use substitution to reduce to finite a.
    [tex]\lim_{x\rightarrow\infty}f(x)=\lim_{x\rightarrow 0}g\left(\frac{1}{x}\right)[/tex]
    where there exist a real number qwt such that f(x)=g(x) for all x>qwt.
    and g(1/0+)=g(1/0-) should either exist, and they either exist of do not together.
     
    Last edited: Aug 22, 2005
  7. Aug 22, 2005 #6

    Hurkyl

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    Staff Emeritus
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    Gold Member

    [tex]\lim_{x\rightarrow\infty}f(x)=\lim_{x\rightarrow 0^+}f\left(\frac{1}{x}\right)[/tex]

    actually. I see that mistake made very often!
     
  8. Aug 22, 2005 #7

    lurflurf

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    Homework Helper

    [tex]\lim_{x\rightarrow\infty}f(x)=\lim_{x\rightarrow 0}f\left(\frac{1}{x}\right)[/tex]

    It is true when both limits exist, perhaps I should have said when the limits exist, or for for suitable f.
     
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