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Math Help - Limits

  1. #1
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    Limits

    Find the limit:
    \lim\limits_{n\to\infty}(\frac{(1+\frac{1}{n})^n}{  e})^n
    the answer is \frac{1}{\sqrt{e}} so i guess we should get something like \lim\limits_{n\to\infty}(1-\frac{1}{2n}).. but dont know what to do with double n and exp. also:
    \lim\limits_{n\to\infty}(\prod\limits_{k=1}^{n}\fr  ac{k}{k+n})^{e^{\frac{1999}{n}}-1} so thats \lim\limits_{n\to\infty}(\frac{n!^2}{2n!})^{e^{\fr  ac{1999}{n}}-1} but again, what to do with exp?
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  2. #2
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    Quote Originally Posted by Julius View Post
    Find the limit:
    \lim\limits_{n\to\infty}(\frac{(1+\frac{1}{n})^n}{  e})^n
    the answer is \frac{1}{\sqrt{e}} so i guess we should get something like \lim\limits_{n\to\infty}(1-\frac{1}{2n}).. but dont know what to do with double n and exp.
    This one is a particular case of the question in this thread.

    lso:
    \lim\limits_{n\to\infty}(\prod\limits_{k=1}^{n}\fr  ac{k}{k+n})^{e^{\frac{1999}{n}}-1} so thats \lim\limits_{n\to\infty}(\frac{n!^2}{2n!})^{e^{\fr  ac{1999}{n}}-1} but again, what to do with exp?
    First of all, start by writing the sequence as \exp((e^{\frac{1999}{n}}-1)\log\prod_{k=1}^n \frac{k}{k+n}). Then, use asymptotic expansions in the exponent. Note that \log\prod_{k=1}\frac{k}{k+n}=\sum_{k=1}^n \log\frac{k/n}{k/n+1}, which evokes a Riemann sum; thus we can avoid Stirling's approximation.
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