The Monstrous Limit

Let $\large\mathfrak L= \displaystyle \lim_{n \to \infty} \left( \dfrac{n!}{n^n} \right)^{1/n}$ $\large\implies \ln \mathfrak L=\displaystyle \lim_{n \to \infty}\dfrac{1}{n} \ln \left( \dfrac{n!}{n^n} \right)$ $\large =\lim_{n \to \infty}\dfrac{1}{n} \ln \left( \dfrac{n\cdot( n-1)\cdot(n-2) \cdots 1}{\underbrace{n\cdot n\cdot n \cdots n}_{\color{#D61F06}{\text{n times}}}} \right)$ $=\lim_{n \to \infty}\dfrac{1}{n} \ln\left( \left(1\right)\cdot \left( 1-\dfrac{1}{n}\right)\cdot \left(1-\dfrac{2}{n}\right)\cdots \left(\dfrac{1}{n}\right)\right)$ $\large =\lim_{n \to \infty}\dfrac{1}{n} \ln \left(\prod_{r=0}^{n- 1}\left(1-\dfrac{r}{n}\right)\right)$ $\large =\lim_{n \to \infty}\dfrac{1}{n} \sum_{r=0}^{n-1}\ln \left(1-\dfrac{r}{n}\right)$ Applying Reimann Sums , $\ln \mathfrak L=\int_0^1 \ln(1-x)\,dx=\int_0^1\ln x\,dx$ $\Large = -1$ $\therefore \mathfrak L=\dfrac 1e\approx \Huge\color{#456461}{\boxed{\color{#EC7300}{\boxed{\color{#0C6AC7}{\mathbf{0.368}}}}}}$

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$\color{#20A900}{\Large{\text{Bonus:}~\lim_{n\to \infty}\dfrac 1n\ln\left(\dfrac{(2n)!}{n^n n!}\right)}}$

This follows from Stirling’s approximation

$n!=\sqrt { 2\pi n } { \left( \dfrac { n }{ e } \right) }^{ n }$

which can be re-written as

$\dfrac { 1 }{ e } ={ \left( \dfrac { n! }{ { n }^{ n } } \right) }^{ \frac { 1 }{ n } }{ \left( 2\pi n \right) }^{ -\frac { 1 }{ 2n } }$

And so the right factor on the right side drops out as $n\rightarrow \infty$

Let's start by considering the following process.

$\displaystyle \lim _{ n\rightarrow \infty }{ { \left( \frac { n! }{ { n }^{ n } } \right) }^{ \frac { 1 }{ n } } } =\lim _{ n\rightarrow \infty }{ { \left( \frac { n! }{ { n }^{ n } } \times \frac { n+1 }{ n+1 } \right) }^{ \frac { 1 }{ n } } }$

$\displaystyle \lim _{ n\rightarrow \infty }{ { \left( \frac { n! }{ { n }^{ n } } \times \frac { n+1 }{ n+1 } \right) }^{ \frac { 1 }{ n } } } =\lim _{ n\rightarrow \infty }{ { \left( \frac { (n+1)! }{ (n+1){ n }^{ n } } \right) }^{ \frac { 1 }{ n } } }$

Let

$\displaystyle n+1=N$

Then

$\displaystyle \lim _{ n\rightarrow \infty }{ { \left( \frac { (n+1)! }{ (n+1){ n }^{ n } } \right) }^{ \frac { 1 }{ n } } } =\lim _{ N\rightarrow \infty }{ { \left( \frac { N! }{ N{ (N-1) }^{ N-1 } } \right) }^{ \frac { 1 }{ n } } }$

$\displaystyle \lim _{ N\rightarrow \infty }{ { \left( \frac { N! }{ { N }^{ N }{ (1-\frac { 1 }{ N } ) }^{ N-1 } } \right) }^{ \frac { 1 }{ n } } } =\lim _{ N\rightarrow \infty }{ { \left( \frac { N!\times e }{ { N }^{ N } } \right) }^{ \frac { 1 }{ n } } }$

Hence,

$\displaystyle \lim _{ n\rightarrow \infty }{ { \left( \frac { n! }{ { n }^{ n } } \right) }^{ \frac { 1 }{ n } } } =\lim _{ n\rightarrow \infty }{ { \left( \frac { (n+1)! \times e}{ { (n+1) }^{ n+1 } } \right) }^{ \frac { 1 }{ n } } }$

And one can deduce that

$\displaystyle \lim _{ n\rightarrow \infty }{ { \left( \frac { n! }{ { n }^{ n } } \right) }^{ \frac { 1 }{ n } } } =\lim _{ n\rightarrow \infty }{ { \left( \frac { (n+k)! \times { e }^{ k }}{ { (n+k) }^{ n+k } } \right) }^{ \frac { 1 }{ n } } }$

Now, let $\displaystyle k=n$

$\displaystyle \lim _{ n\rightarrow \infty }{ { \left( \frac { n! }{ { n }^{ n } } \right) }^{ \frac { 1 }{ n } } } =\lim _{ n\rightarrow \infty }{ { \left( \frac { (n+n)!\times { e }^{ n } }{ { (n+n) }^{ n+n } } \right) }^{ \frac { 1 }{ n } } }$

$\displaystyle \lim _{ n\rightarrow \infty }{ { \left( \frac { (2n)! \times { e }^{ n } }{ { (2n) }^{ 2n }} \right) }^{ \frac { 2 }{ 2n } } } ={ \left( \lim _{ n\rightarrow \infty }{ { \left( \frac { (2n)! }{ { (2n) }^{ 2n } } \right) }^{ \frac { 1 }{ 2n } } } \times { e }^{ \frac { 1 }{ 2 } } \right) }^{ 2 }$

Let the limit equal to L then,

$\displaystyle { \left( \lim _{ n\rightarrow \infty }{ { \left( \frac { (2n)! }{ { (2n) }^{ 2n } } \right) }^{ \frac { 1 }{ 2n } } } \times { e }^{ \frac { 1 }{ 2 } } \right) }^{ 2 }={ L }^{ 2 }e=L$

$\displaystyle L=\boxed{\frac { 1 }{ e }}$