A Stirling result

$\lim_{n \to \infty} \dfrac{\ln \frac{n^n}{n!}}{n}$

= $\dfrac{1}{n} \displaystyle \sum_{r=0}^{n-1} \ln \bigg(\dfrac{n}{n - r}\bigg)$

= $\dfrac{1}{n} \displaystyle \sum_{r=0}^{n-1} \ln \bigg(\dfrac{1}{1-r/n}\bigg)$

= $-\displaystyle \int_{0}^{1} \ln(1-x) {\mathrm dx}$

= $1$

By Stirling's approximation ,

$\displaystyle \lim_{x \rightarrow \infty} \frac{\ln \frac{x^{x}}{x!}}{x} = \displaystyle \lim_{x \rightarrow \infty} \frac{\ln \frac{e^{x}}{\sqrt{2 \pi x}}}{x} = \displaystyle \lim_{x \rightarrow \infty} \frac{x-\ln \sqrt{2 \pi x}}{x} = \boxed{1}$

First of all, credit to a guy named "Matthew S." for helping me. I'm not sure if he's on this site.

$\lim_{x\rightarrow \infty} \frac{\ln\frac{x^x}{x!}}{x}\\=\frac{ \infty}{ \infty}$

Use l'Hôpital's rule:

$\lim_{x\rightarrow \infty} \frac{\frac{d}{dx} (\ln\frac{x^x}{x!})}{\frac{d}{dx}(x)} \\ =\lim_{x\rightarrow \infty} \frac{d}{dx} (\ln\frac{x^x}{x!}) \\ =\lim_{x\rightarrow \infty} \frac{d}{dx} (\ln x^x - \ln x!) \\ =\lim_{x\rightarrow \infty} \frac{d}{dx} [x\ln x-(\ln x+\ln (x-1) + \ln (x-2) + \dots+ \ln (2) + \ln (1))] \\ =\lim_{x\rightarrow \infty} \ln x + 1 - (\frac{1}{x} + \frac{1}{x-1} + \frac{1}{x-2} + \dots + 0 + 0)$ $\frac{1}{x} + \frac{1}{x-1} + \frac{1}{x-2} + \dots + 0 + 0 \approx \ln x+\gamma+\epsilon_{0}$

$=\lim_{x\rightarrow \infty} \ln x + 1 - \ln x+\gamma+\epsilon_{0} \\ =1+\gamma+\epsilon_{0}$

The error, $\gamma$ and $\epsilon_{0}$ , goes to $0$ with more terms, so just

$\boxed{1}$

Can be solved easily using limit as a sum concept.

Above written question is summation of (1/x)*ln(x/r) with r varying from 1 to x.

Replacing 1/x with dn and r/x with n wet get ((integral of 0 to 1) ln(1/n))dn which on solving yields 1 as answer.

Please reply if you don't know why I replaced 1/x with dn and r/x with n.

When $x\rightarrow \infty$ , the expression $\ln\left( \frac { { x }^{ x } }{ x! } \right)$ also tends to infinity, because ${ x }^{ x }$ grows faster than $x!$ . Then, the limit has to be calculated by looking at the growing rates of the nominator and the denominator.

Hypothesis: $\frac { { x }^{ x } }{ x! }$ grows as fast as ${a}^{x}$ , (where $a$ is a constant) when $x \rightarrow \infty$ . Therefore:

$\lim _{ x\rightarrow \infty }{ \frac { \ln { \left( \frac { { x }^{ x } }{ x! } \right) } }{ x } } =\lim _{ x\rightarrow \infty }{ \frac { \ln { \left( { a }^{ x } \right) } }{ x } } =\lim _{ x\rightarrow \infty }{ \frac { x\cdot \ln { \left( a \right) } }{ x } } =\ln { \left( a \right) }$

If the above hypothesis is true, then each term of the sequence determined by $\frac { { x }^{ x } }{ x! }$ has to be equal to the previous term times the constant $a$ when $x\rightarrow \infty$ . That is:

$\frac { { (x+1) }^{ (x+1) } }{ (x+1)! } =a\frac { { x }^{ x } }{ x! } \\ a=\frac { { (x+1) }^{ (x+1) } }{ { x }^{ x } } \cdot \frac { x! }{ (x+1)! } \\ a=\frac { { (x+1) }^{ (x+1) } }{ (x+1)\cdot { x }^{ x } } \\ a=\frac { { (x+1) }^{ x } }{ { x }^{ x } } \\ a={ \left( 1+\frac { 1 }{ x } \right) }^{ x }$

If $x\rightarrow \infty$ , then we see that value of $a$ is obtained by the fundamental limit:

$a=\lim _{ x\rightarrow \infty }{ { \left( 1+\frac { 1 }{ x } \right) }^{ x } } =e$

Finally:

$\lim _{ x\rightarrow \infty }{ \frac { \ln { \left( \frac { { x }^{ x } }{ x! } \right) } }{ x } } =\ln { \left( a \right) } =\ln { \left( e \right) =1 }$

As x tends to infinity the following facts are worth noting; 1:x^x will reach infinity quicker than x! and is therefore infinite times bigger than x! (this because its a super exponential function which grows faster than a factorial one) 2:log of infinity is still infinity and 3:infinity/infinity is 1. As a result the numerator will become log(infinity) which is still infinity and the denominator is also infinity therefore the answer is 1. (and NO I did not even touch Wolfram Alpha or Mathematica for that matter on this occasion).

First some facts

$x^x > x!$

$\frac {x}{x} = 1$

$In\infty = \infty$

So let's put in $\infty$ as our $x$

$\displaystyle\lim_{x\rightarrow\infty} \frac {In\frac{\infty^\infty}{\infty!}}{\infty}$

Since $\infty^\infty$ is $\infty$ times larger than $\infty!$ we can say that $\frac {\infty^\infty}{\infty!} = \infty$ .

$\frac{In\infty}{\infty}$

We also know that $In\infty = \infty$ so

$\frac {\infty}{\infty} = 1$

So the limit is $1$