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Evaluation of $\displaystyle \lim_{n\rightarrow \infty}\frac{n!\cdot e^n}{\sqrt{n}\cdot n^n}$

$\bf{My\; Try::}$we can write it as $$l=\lim_{n\rightarrow \infty}\frac{e^n}{\sqrt{n}}\cdot \left(\frac{1}{n}\cdot \frac{2}{n}\cdot \frac{3}{n}\cdots \cdots \frac{n}{n}\right)$$

$$\ln (l) = \lim_{n\rightarrow \infty}\bigg[n-\frac{1}{2}n+\sum^{n}_{r=1}\ln\left(\frac{r}{n}\right)\bigg]$$

Now how can i solve it, Help required, Thanks

  • 0
    That last summation is looking like it's begging to become a riemann sum (just a guess, might not work) – 2017-01-27
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    ...you could also use the limit-product form of the factorial/gamma function. – 2017-01-27
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    the result should be $$\sqrt{2\pi}$$ – 2017-01-27
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    To Simply Beautiful plz explain here, thanks. Yes Dr. you are right. plz explain – 2017-01-27
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    http://en.wikipedia.org/wiki/Gamma_function#Euler.27s_definition_as_an_infinite_product – 2017-01-27
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    If you find the limit, you are proving Stirling. – 2017-01-27
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    Your calculation of $\ln(l)$ is incorrect, incidentally - you correctly take the log of $e^n$ as $n$, but then you take the log of $\sqrt{n}$ as $\frac12n$... – 2017-01-27
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    @StevenStadnicki you right, I removed it. – 2017-01-27
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    @zhw.'s comment makes that I fail to understand the question. At present it seems to ask for a proof of Stirling's equivalent. Is that what you want to ask? – 2017-01-27
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    Equivalent to proving the Wallis infinite product $\pi=4\frac {2^2}{1\cdot 3} \frac {4^2}{3\cdot 5} \frac {6^2}{5\cdot 7}...$ – 2017-01-27
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    @user254665 I'd like to see an answer based on this observation. – 2017-01-27

1 Answers 1

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Sorry, your question is not clear. What do you mean "without Stirling approximation" ? Because you use Stirling's construction.

$\displaystyle c:=\lim_{n\to\infty}\frac{n!e^n}{\sqrt{n} n^n}$

$\displaystyle =>\enspace \lim_{n\to\infty}\frac{(2n)!e^{2n}}{\sqrt{2n} (2n)^{2n}}=c ~~, ~~~\lim_{n\to\infty}\frac{n!^2 e^{2n}}{\sqrt{n}^2 n^{2n}}=c^2$

We devide the second by the first, means $c^2/c$ :

$\displaystyle =>\enspace c=\sqrt{2}\lim_{n\to\infty}\frac{n!^2 2^{2n} }{ \sqrt{n} (2n)! }=\sqrt{2}~\Gamma\left(\frac{1}{2}\right)=\sqrt{2\pi} $

If the Gamma function is not enough then look at the Wallis product to get the value of $~c~$ .