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Thread: Strange limit question

  1. #1
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    Strange limit question

    Hi All,

    I want to know whether there is an easy way to see how fast

    $\displaystyle \lim_{n \to \infty}(1 - (1-1/n^a)^n)$

    goes to zero as a function of $\displaystyle a~(a > 1)$.

    I already know (thanks Maple!) that for $\displaystyle 1 \le a<2$ the function goes slower than $\displaystyle n^{-1}$, i.e.

    $\displaystyle \lim_{n \to \infty}n (1 - (1-1/n^a)^n) = \infty~\text{for}~1 \le a < 2$

    whereas

    $\displaystyle \lim_{n \to \infty}n (1 - (1-1/n^a)^n) = 0~\text{for}~a>2 $.

    (and for $\displaystyle a=2$ the limit is 1). But is there a clear and easy way to see this? I can't even do the limit by hand because it seems to require an infinite number of applications of L'hopitals rule...

    Note: Obviously the case $\displaystyle a=1$ is easy since I can substitute $\displaystyle e^{-1}$ and the thing goes to a constant .. but what about the cases when it really goes to zero? Obviously a=2 is special.

    Thanks in advance.
    Last edited by galois; May 20th 2010 at 11:00 AM.
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  2. #2
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    Ah, OK, nevermind, I think I see it using the binomial formula on the $\displaystyle (1 - (1-1/n^a)^n)$ term, giving a sum where one can deduce the stated property by factorization (the 1 and the first term in the series cancel).
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  3. #3
    MHF Contributor

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    You might also note that $\displaystyle \lim_{n\to\infty} (1+ \frac{1}{n})^n$ is commonly used as a definition of e. If we replace the "n" in $\displaystyle (1- \frac{1}{n^a})^n$ with $\displaystyle n^{1/a}$ (assuming, for the moment, that $\displaystyle a\ne 0$), we get $\displaystyle (1- \frac{1}{n})^{n/a}$. Now replace that n by "-n" to get $\displaystyle (1+ \frac{1}{n})^{-n/a}= [(1+ \frac{1}{n})^n]^{-1/a}= e^{-1/a}= \frac{1}{e^{1/a}}$
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