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Math Help - [SOLVED] Pointwise and Uniform Convergence of x^n e^{-nx}

  1. #1
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    [SOLVED] Pointwise and Uniform Convergence of x^n e^{-nx}

    There is a thread similar to this already in existence, but it is quite old and it didn't help me very much.
    I need to determine the convergence of
     \lim_{n \to \infty} x^n e^{-nx}
    for x \in [0,1] and x \in [0, +\infty)
    Trying to use l'Hopital's Rule I get
     \lim_{n \to \infty} x^n e^{-nx} = \lim_{n \to \infty} \frac{x^n}{e^{nx}}
     \lim_{n \to \infty} \frac{x^n}{e^{nx}} \stackrel{H}{=} \frac{\log(x) x^{n-1}}{e^{nx}}
    Using l'Hopital n time results in
     \lim_{n \to \infty} \frac{x^n}{e^{nx}} \stackrel{H}{=} \frac{(\log x)^n}{e^{nx}}
    However, I am not sure this even makes sense, because f_n is not continuous with respect to n. So, what am I suppose to do?
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  2. #2
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    Quote Originally Posted by lvleph View Post
    There is a thread similar to this already in existence, but it is quite old and it didn't help me very much.
    I need to determine the convergence of
     \lim_{n \to \infty} x^n e^{-nx}
    for x \in [0,1] and x \in [0, +\infty)
    Trying to use l'Hopital's Rule I get
     \lim_{n \to \infty} x^n e^{-nx} = \lim_{n \to \infty} \frac{x^n}{e^{nx}}
     \lim_{n \to \infty} \frac{x^n}{e^{nx}} \stackrel{H}{=} \frac{\log(x) x^{n-1}}{e^{nx}}
    Using l'Hopital n time results in
     \lim_{n \to \infty} \frac{x^n}{e^{nx}} \stackrel{H}{=} \frac{(\log x)^n}{e^{nx}}
    However, I am not sure this even makes sense, because f_n is not continuous with respect to n. So, what am I suppose to do?
    Is this a series or a sequence?
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  3. #3
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    It is a sequence.

    What I have to do is determine Piecewise and Uniform convergence on those intervals. I can get the Uniform once I get the Piecewise convergence. But, I can't get past that.
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  4. #4
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    Quote Originally Posted by lvleph View Post
    There is a thread similar to this already in existence, but it is quite old and it didn't help me very much.
    I need to determine the convergence of
     \lim_{n \to \infty} x^n e^{-nx}
    for x \in [0,1] and x \in [0, +\infty)
    Trying to use l'Hopital's Rule I get
     \lim_{n \to \infty} x^n e^{-nx} = \lim_{n \to \infty} \frac{x^n}{e^{nx}}
     \lim_{n \to \infty} \frac{x^n}{e^{nx}} \stackrel{H}{=} \frac{\log(x) x^{n-1}}{e^{nx}}
    Using l'Hopital n time results in
     \lim_{n \to \infty} \frac{x^n}{e^{nx}} \stackrel{H}{=} \frac{(\log x)^n}{e^{nx}}
    However, I am not sure this even makes sense, because f_n is not continuous with respect to n. So, what am I suppose to do?
    It is clearly pointwise convergent. Now, do you think it's uniformly convergent? Will your choice of N in the definition of convergence depend on x?
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  5. #5
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    Like I said I don't need help with the Uniform Convergence. If it is clearly Pointwise then it should be easy to prove, however I am unable to figure out how.
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  6. #6
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    So, I have a bit of an idea, but can't seem to figure it out.
    \lim_{n \to \infty} \left(\frac{x}{e^x}\right)^n
    This will converge to zero if \left(\frac{x}{e^x}\right)^n < 1 and will converge to one if \left(\frac{x}{e^x}\right)^n = 1, but diverges if \left(\frac{x}{e^x}\right)^n > 1
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  7. #7
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    Consider the function f(x)=\ln (x) -x then f'(x)= \frac{1}{x} -1 and thus f is increasing on (0,1) and decreasing on (1, \infty) and so f(x)\leq f(1)=-1 for all x\in (0, \infty). Now consider g_y(x)= x^ye^{-yx}=e^{y(\ln (x) -x)} with y>0 then 0\leq |g_y(x)| \leq |e^{-y} | \rightarrow 0 as y\rightarrow \infty. This proves both pointwise and uniform convergence in [0,\infty )
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