Thread: Limits

Find the limit:
$\displaystyle \lim\limits_{n\to\infty}(\frac{(1+\frac{1}{n})^n}{ e})^n$
the answer is $\displaystyle \frac{1}{\sqrt{e}}$ so i guess we should get something like $\displaystyle \lim\limits_{n\to\infty}(1-\frac{1}{2n})$.. but dont know what to do with double n and exp. also:
$\displaystyle \lim\limits_{n\to\infty}(\prod\limits_{k=1}^{n}\fr ac{k}{k+n})^{e^{\frac{1999}{n}}-1}$ so thats $\displaystyle \lim\limits_{n\to\infty}(\frac{n!^2}{2n!})^{e^{\fr ac{1999}{n}}-1}$ but again, what to do with exp?

Originally Posted by Julius

Find the limit:
$\displaystyle \lim\limits_{n\to\infty}(\frac{(1+\frac{1}{n})^n}{ e})^n$
the answer is $\displaystyle \frac{1}{\sqrt{e}}$ so i guess we should get something like $\displaystyle \lim\limits_{n\to\infty}(1-\frac{1}{2n})$.. but dont know what to do with double n and exp.

This one is a particular case of the question in this thread.

lso:
$\displaystyle \lim\limits_{n\to\infty}(\prod\limits_{k=1}^{n}\fr ac{k}{k+n})^{e^{\frac{1999}{n}}-1}$ so thats $\displaystyle \lim\limits_{n\to\infty}(\frac{n!^2}{2n!})^{e^{\fr ac{1999}{n}}-1}$ but again, what to do with exp?

First of all, start by writing the sequence as $\displaystyle \exp((e^{\frac{1999}{n}}-1)\log\prod_{k=1}^n \frac{k}{k+n})$. Then, use asymptotic expansions in the exponent. Note that $\displaystyle \log\prod_{k=1}\frac{k}{k+n}=\sum_{k=1}^n \log\frac{k/n}{k/n+1}$, which evokes a Riemann sum; thus we can avoid Stirling's approximation.