Evaluation of $\lim_{n\rightarrow \infty}\frac{n!\cdot e^n}{\sqrt{n}n^n}$ without stirling approximation

Question

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

That last summation is looking like it's begging to become a riemann sum (just a guess, might not work) — ASKASK, Jan 27 '17 at 18:47
...you could also use the limit-product form of the factorial/gamma function. — Simply Beautiful Art, Jan 27 '17 at 18:49
To Simply Beautiful plz explain here, thanks. Yes Dr. you are right. plz explain — juantheron, Jan 27 '17 at 18:56
http://en.wikipedia.org/wiki/Gamma_function#Euler.27s_definition_as_an_infinite_product — Simply Beautiful Art, Jan 27 '17 at 19:11
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 $\frac12n$... — Steven Stadnicki, Jan 27 '17 at 19:46
@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? — Did, Jan 27 '17 at 20:01
Equivalent to proving the Wallis infinite product $\pi=4\frac {2^2}{1\cdot 3} \frac {4^2}{3\cdot 5} \frac {6^2}{5\cdot 7}...$ — DanielWainfleet, Jan 27 '17 at 20:09
@user254665 I'd like to see an answer based on this observation. — Rene Schipperus, Jan 27 '17 at 20:47
@juantheron : Is my answer to your question sufficient or do you want a different one? If yes, which one? — user90369, Apr 25 '19 at 15:16

Gary · Accepted Answer · 2024-03-28T07:13:14.073

Let $$ a_n : = \frac{{n!\,\mathrm{e}^n }}{{n^n \sqrt n }} $$ and $a:= \lim a_n$. I assume that this limit exists and is finite and positive. Then \begin{align*} a_{2n} & = \frac{{(2n)!\,\mathrm{e}^{2n} }}{{(2n)^{2n} \sqrt {2n} }} = \frac{{(2n)!}}{{n!^2 2^{2n} }}\frac{{(n!)^2 \mathrm{e}^{2n} }}{{n^{2n + 1} }}\sqrt {\frac{n}{2}} = \sqrt {\frac{{1 \cdot 1}}{{2 \cdot 2}}\frac{{3 \cdot 3}}{{4 \cdot 4}} \cdots \frac{{(2n - 1)(2n - 1)}}{{2n \cdot 2n}}} a_n^2 \sqrt {\frac{n}{2}} \\ & = \sqrt {\frac{{1 \cdot 3}}{{2 \cdot 2}}\frac{{3 \cdot 5}}{{4 \cdot 4}} \cdots \frac{{(2n - 1)(2n + 1)}}{{2n \cdot 2n}}} a_n^2 \sqrt {\frac{n}{{4n + 2}}} . \end{align*} Taking the limit of both sides gives $$ a = \sqrt {\mathop {\lim }\limits_{n \to + \infty } \frac{{1 \cdot 3}}{{2 \cdot 2}}\frac{{3 \cdot 5}}{{4 \cdot 4}} \cdots \frac{{(2n - 1)(2n + 1)}}{{2n \cdot 2n}}} a^2 \frac{1}{2}. $$ The infinite product under the square root is the reciprocal of the famous Wallis product, whence $$ a = \sqrt {\frac{2}{\pi }} a^2 \frac{1}{2}, $$ i.e., $a=\sqrt{2\pi}$.

score 1 · Answer 2 · edited Mar 26 '21 at 19:45

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

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

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

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

$\displaystyle \implies \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 $~c~$ .

Evaluation of $\lim_{n\rightarrow \infty}\frac{n!\cdot e^n}{\sqrt{n}n^n}$ without stirling approximation

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