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Evaluating limit at infinity by Maclaurin series

  1. Apr 10, 2017 #1
    1. The problem statement, all variables and given/known data
    I've begun going through Boas' Math Methods in the Physical Sciences and am stuck on problem 1.15.25. The problem is to evaluate
    ## \lim_{x\to \infty } x^n e^{-x} ##
    By using the Maclaurin expansion for ##e^{x}##.

    2. Relevant equations
    We know the Maclaurin expansion for the exponential function to be
    ## 1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\text{...}##
    We are also given the hint to divide the numerator and denominator by ##x^n## before taking the limit.

    3. The attempt at a solution
    Taking the hint I proceed
    ##
    e^{-x} x^n=\frac{x^n}{e^x}=\frac{1}{\frac{e^x}{x^n}}=\frac{1}{\frac{1}{x^n}\left(1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\text{...}\right)}=\frac{1}{\frac{1}{x^n}+\frac{x}{x^n}+\frac{x^2}{2 x^n}+\frac{x^3}{6 x^n}+\frac{x^4}{24 x^n}+\text{...}}=\frac{1}{\frac{1}{x^n}+\frac{1}{x^{n-1}}+\frac{1}{2 x^{n-2}}+\frac{1}{6 x^{n-3}}+\frac{1}{24 x^{n-4}}+\text{...}}

    ##
    I really don't see how this helps me as I let x go to infinity.
    Any suggestions?
     
  2. jcsd
  3. Apr 10, 2017 #2

    stevendaryl

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    I had Ralph Boas (husband of the author of that book) for a math course once. He was a very sweet old guy.

    I'm not exactly sure how she wants you to solve the problem, but if you could prove that, for [itex]x[/itex] large enough, [itex]e^x > x^{n+1}[/itex], then it would follow that [itex]x^n/e^x < 1/x[/itex].
     
  4. Apr 10, 2017 #3

    ehild

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    The denominator is ##\frac{1}{x^n}+\frac{1}{x^{n-1}}+\frac{1}{2 x^{n-2}}+\text{...}+\frac{1}{n!}+\frac{1}{(n+1)!}x+\frac{1}{(n+2)!}x^2+\text{...}##
    What happens if x goes to infinity?
     
  5. Apr 12, 2017 #4
    As x goes to infinity, ##\frac{1}{x^n}, \frac{1}{x^{n-1}}, \frac{1}{2 x^{n-2}},...## go to zero leaving us to evaluate the rest of the series ##\frac{1}{n!}+\frac{1}{(n+1)!}x+\frac{1}{(n+2)!}x^2+\text{...}##

    The only step I can see from here is to factor out an n!.
    ##\frac{1}{n!}+\frac{1}{(n+1)!}x+\frac{1}{(n+2)!}x^2+\text{...} = \frac{1}{n!}+\frac{x}{(n+1) n!}+\frac{x^2}{(n+2) (n+1) n!}+\text{...} = \frac{1}{n!}\left(1+\frac{x}{n+1}+\frac{x^2}{(n+2) (n+1)}+\text{...}\right)##
    And once again I'm stuck as I know of no way to evaluate this.
     
  6. Apr 12, 2017 #5

    Ray Vickson

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    For ##x > 0##, all the terms in the series for ##e^x## are positive, so ##e^x > x^{n+k}/(n+k)!, ## hence
    $$\frac{x^n}{e^x} < \frac{x^n}{x^{n+k}/(n+k)!} = (n+k)!/x^k. $$
     
  7. Apr 12, 2017 #6

    vela

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    Seems like you're making this more complicated than it needs to be. Doesn't
    $$\lim_{x\to\infty} \frac{1}{n!} + \frac{1}{(n+1)!} x + \cdots$$ diverge?
     
  8. Apr 12, 2017 #7

    ehild

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    Remember that n is a finite number.
    ##1+\frac{x}{n+1}+\frac{x^2}{(n+2) (n+1)}+\text{...}>1+\frac{x}{n+1}## which goes to infinity if x-->∞. What is the limit of its reciprocal?
     
  9. Apr 12, 2017 #8
    Thank you all! The fact that n is a fixed finite constant was the fundamental concept I wasn't using in my solution. I now can see that once I take the limit of the reciprocal sum it goes to zero =)

    You have no idea how frustrating this problem has been (for no good reason). Thank you al for sharing your insights with a fresh pair of eyes!
     
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