## Lemma 1: $\lim_{n \rightarrow \infty} \sqrt[n]{n!}/n=1/e$

### Way 1 (somewhat rigorous)

From elementary calculus, we have that: $\int_{0}^{1} \ln(x) dx =-1$ Taking this as a Riemann sum, as done in introductory calculus, we have: $-1=\int_{0}^{1}\ln(x)dx=\lim_{N \rightarrow \infty} \sum_{k=1}^{N}\ln\left (\frac{k}{N} \right ) \cdot \frac{1}{N}$ $-1=\lim_{N \rightarrow \infty} -\ln(N)+\frac{1}{N} \sum_{k=1}^{N}\ln\left (k \right )$ $-1=\lim_{N \rightarrow \infty} -\ln(N)+\frac{1}{N} \ln\left (N! \right )$ Therefore, $\lim_{N \rightarrow \infty} \frac{\sqrt[N]{N!}}{N}=\frac{1}{e}$

### Way 2 (less rigorous)

$\lim_{n \rightarrow \infty} \frac{\sqrt[n]{n!}}{n}=x$ So, for n big, in a certain sense: $n! \approx (nx)^n$ $\frac{(n+1)!}{n!(n+1)}=1 \approx \frac{((n+1)x)^{n+1}}{(nx)^n (n+1)}=\left ( 1+ \frac{1}{n} \right )^n x$ Thus, in order to get equality in the limit, we must have: $x = \lim_{n \rightarrow \infty} \left ( 1+ \frac{1}{n} \right )^{-n}=\frac{1}{e}$

## Lemma 2: Wallis Product in Factorial Form

Recall from this article the following expression for pi: $\frac{\pi}{2}=\prod_{k=1}^{\infty}\frac{2k \cdot 2k}{(2k-1)(2k+1)}=\lim_{N \rightarrow \infty}\prod_{k=1}^{N}\frac{2k \cdot 2k}{(2k-1)(2k+1)}=\lim_{N \rightarrow \infty} \frac{\left ( 2^N \cdot N! \right )^4}{\left ( (2N)! \right )^2(2N+1)}$

## Lemma 3: An Inequality for the Natural Logarithm

Let $x,y > 0$. Clearly $0 \leq \frac{1}{y^2 (1+y)^2 (2y+1)^2}$ Therefore $0 \leq \int_{x}^{\infty}\frac{dy}{y^2 (1+y)^2 (2y+1)^2}=\frac{1}{x}+\frac{1}{x+1}+\frac{4}{x+1/2}-6\ln \left ( 1+\frac{1}{x} \right )$ $6\ln \left ( 1+\frac{1}{x} \right ) -\frac{6}{x+1/2} \leq \frac{1}{x}+\frac{1}{x+1}-\frac{2}{x+1/2}$ $(x+\tfrac{1}{2})\ln \left ( 1+\frac{1}{x} \right ) -1 \leq \frac{(x+\tfrac{1}{2})}{6}\left (\frac{1}{x}+\frac{1}{x+1} \right )-\frac{1}{3}=\frac{1}{12x(x+1)}$ Also, clearly $0 \leq \frac{16y^2+41y+24}{y(1+y)^2 (2+y)^2 (2y+1)^2}$ Therefore $0 \leq \int_{x}^{\infty}\frac{16y^2+41y+24}{y(1+y)^2 (2+y)^2 (2y+1)^2} dy=6\left (\ln \left ( 1+\frac{1}{x} \right )-\frac{1}{x+\tfrac{1}{2}} \right)-\frac{1}{2(x+\tfrac{1}{2})(x+1)(x+2)}$ And so $\frac{1}{12(x+1)(x+2)} \leq (x+\tfrac{1}{2})\ln \left ( 1+\frac{1}{x} \right )-1$

## Theorem: Stirling's Approximation

Let us define a function and sequence of coefficients as follows: $g(n)=\ln\left ( \frac{n!}{\left ( \tfrac{n}{e} \right )^n \sqrt{2\pi n}} \right )=\sum_{k=-\infty}^{\infty} A_k n^k$ We then have, from lemma 1, $\frac{1}{e}=\lim_{n \rightarrow \infty} \frac{\sqrt[n]{n!}}{n}=\lim_{n \rightarrow \infty} \frac{\sqrt[n]{\left (\tfrac{n}{e} \right )^n \sqrt{2\pi n} \cdot e^{g(n)}}}{n}=\frac{1}{e} \lim_{n \rightarrow \infty} \sqrt[2n]{2\pi n} \cdot e^{g(n)/n}$ Thus $1=\lim_{n \rightarrow \infty} e^{g(n)/n}=\exp\left (\lim_{n \rightarrow \infty} \sum_{k=-\infty}^{\infty} A_k n^{k-1} \right )=\exp\left (\lim_{n \rightarrow \infty} \sum_{k=1}^{\infty} A_k n^{k-1} \right )$ And therefore $A_k=0$ for $k \geq 1$. From lemma 2, $\frac{\pi}{2}=\lim_{n \rightarrow \infty} \frac{\left ( 2^n \cdot n! \right )^4}{\left ( (2n)! \right )^2(2n+1)}=\lim_{n \rightarrow \infty} \frac{\left ( 2^n \cdot \left (\tfrac{n}{e} \right )^n \sqrt{2\pi n} \cdot e^{g(n)} \right )^4}{\left ( \left (\tfrac{2n}{e} \right )^{2n} \sqrt{4\pi n} \cdot e^{g(2n)} \right )^2(2n+1)}$ $\frac{\pi}{2}=\lim_{n \rightarrow \infty} \frac{\left ( 2^n \cdot n! \right )^4}{\left ( (2n)! \right )^2(2n+1)}=\lim_{n \rightarrow \infty} \frac{\left ( 2^n \cdot \left (\tfrac{n}{e} \right )^n \sqrt{2\pi n} \cdot e^{g(n)} \right )^4}{\left ( \left (\tfrac{2n}{e} \right )^{2n} \sqrt{4\pi n} \cdot e^{g(2n)} \right )^2(2n+1)}$ $\frac{\pi}{2}=\lim_{n \rightarrow \infty} \frac{n \pi}{2n+1} \cdot e^{4g(n)-2g(2n)}$ $0=\lim_{n \rightarrow \infty} 2g(n)-g(2n)=2A_0-A_0=A_0$ Therefore, $A_k=0$ for $k \geq 0$, and thus $\lim_{n \rightarrow \infty} g(n)=0$. Thus it follows that $\lim_{n \rightarrow \infty} \frac{n!}{\left ( \tfrac{n}{e} \right )^n \sqrt{2\pi n}}=1$ This fact is known as Stirling's Approximation. Moreover, we have $g(n)-g(n+1)=\ln\left ( \frac{n!\left ( \tfrac{n+1}{e} \right )^{n+1} \sqrt{2\pi (n+1)}}{(n+1)!\left ( \tfrac{n}{e} \right )^n \sqrt{2\pi n}} \right )=\ln\left ( \frac{(n+1)^{n+\tfrac{1}{2}}}{e \cdot n^{n+\tfrac{1}{2}}} \right )$ $g(n)-g(n+1)=(n+\tfrac{1}{2})\ln\left ( 1+\frac{1}{n} \right )-1$ By lemma 3, we then have $\frac{1}{12(n+1)(n+2)} \leq g(n)-g(n+1) \leq \frac{1}{12n(n+1)}$ $\sum_{k=n}^{\infty} \frac{1}{12(k+1)(k+2)}=\frac{1}{12(n+1)} \leq \sum_{k=n}^{\infty} g(k)-g(k+1)=g(n)-g(\infty)=g(n) \leq \sum_{k=n}^{\infty} \frac{1}{12k(k+1)}=\frac{1}{12n}$ That is $\tfrac{1}{12(n+1)} \leq g(n) \leq \tfrac{1}{12n}$. And therefore: $\left (\tfrac{n}{e} \right )^n \sqrt{2\pi n}\cdot e^{\tfrac{1}{12(n+1)}} \leq n! \leq \left (\tfrac{n}{e} \right )^n \sqrt{2\pi n} \cdot e^{\tfrac{1}{12n}}$ In fact, it is possible to obtain exact formulas for $g(n)$. For example, by more advanced calculations, we can show that $g(n)=\int_{0}^{\infty}\frac{2 \tan^{-1}\left ( \tfrac{y}{n} \right )}{e^{2\pi y}-1}=\sum_{k=1}^{\infty} \frac{B_{2k}}{2k(2k-1)n^{2k-1}}=\frac{1}{12n}-\frac{1}{360n^3}+\frac{1}{1260n^5}- \cdots$ Where $B_m$ is the mth Bernoulli number. These two expressions are, respectively Binet's second expression and Stirling's series.

## Corollary: Product of a Rational Function

Firstly, since $\prod_{k=1}^N \left(ak+b\right)=a^N\prod_{k=1}^N \left(k+\frac{b}{a}\right)$ We will just evaluate $\prod_{k=1}^N \left(k+b\right)=\frac{(N+b)!}{b!} \approx \left(\frac{N+b}{e}\right)^{N+b}\frac{\sqrt{2\pi(N+b)}}{b!}=N^{N+b+\tfrac{1}{2}}e^{-N}\frac{\sqrt{2\pi}}{b!}e^{-b}\left(1+\frac{b}{N}\right)^{N+b+\tfrac{1}{2}}$ $\prod_{k=1}^N \left(k+b\right)=\frac{(N+b)!}{b!} \approx N^{N+b+\tfrac{1}{2}}e^{-N}\frac{\sqrt{2\pi}}{b!}$ More generally, given the above, it is not difficult to demonstrate the following generalization. Let $m,n > 0$. Let $a_1,a_2,...,a_m$ and $b_1,b_2,...,b_n$ and $r_1,r_2,...,r_m$ and $s_1,s_2,...,s_n$ be sequences of numbers, such that $\sum_{k=1}^m r_k=\sum_{k=1}^n s_k$ and $\sum_{k=1}^m a_k r_k=\sum_{k=1}^n b_k s_k$ Then $\prod_{k=1}^\infty\frac{(k+a_1)^{r_1}(k+a_2)^{r_2}\cdots (k+a_m)^{r_m}}{(k+b_1)^{s_1}(k+b_2)^{s_2}\cdots (k+b_n)^{s_n}}=\frac{\prod_{j=1}^n (b_j!)^{s_j}}{\prod_{j=1}^m (a_j!)^{r_j}}$ In cases where the coefficients are non-integral, we use the Gamma function (an extension of the factorial to non-integers), instead of factorials: $\prod_{k=1}^\infty\frac{(k+a_1)^{r_1}(k+a_2)^{r_2}\cdots (k+a_m)^{r_m}}{(k+b_1)^{s_1}(k+b_2)^{s_2}\cdots (k+b_n)^{s_n}}=\frac{\prod_{j=1}^n (\Gamma (b_j+1))^{s_j}}{\prod_{j=1}^m (\Gamma (a_j+1))^{r_j}}$ For instance $\prod_{k=0}^\infty \frac{(k+1)(k+a+b)}{(k+a)(k+b)}=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}=B(a,b)$ $\frac{\sin(\pi x)}{\pi x}=\prod_{k=1}^\infty\frac{(k-x)(k+x)}{k^2}=\frac{\Gamma(1)^2}{\Gamma(1-x) \Gamma(1+x)}=\frac{1}{\Gamma(1-x)x \Gamma(x)}$ $\prod_{k=1}^\infty\frac{(1+\tfrac{1}{k})^x}{1+\tfrac{x}{k}}=\prod_{k=1}^\infty\frac{(k+1)^x k}{k^x (k+x)}=\frac{\Gamma(1)^x \Gamma (1+x)}{\Gamma(2)^x \Gamma(1)}=\Gamma(1+x)$

## Corollary: Asymptotic Behavior of Bernoulli Numbers

In this article, we found that $\zeta(2n)=\frac{1}{2}\frac{(2\pi)^{2n}}{(2n)!}\left | B_{2n} \right |$ Combining this with Stirling's approximation, we find that $\left | B_{2n} \right |=2\zeta(2n)\frac{(2n)!}{(2\pi)^{2n}} \approx 4\left ( \frac{n}{\pi e} \right )^{2n} \sqrt{n\pi} \cdot e^{1/24n}$

## Corollary: Approximation for Binomial Coefficients

$\binom{a}{b}=\frac{a!}{b!(a-b)!} \approx \frac{\left (\tfrac{a}{e} \right )^a \sqrt{2\pi a}} {\left (\tfrac{b}{e} \right )^b \sqrt{2\pi b}\left (\tfrac{a-b}{e} \right )^{a-b} \sqrt{2\pi (a-b)}}=\frac{1}{\sqrt{2\pi}}\sqrt{\frac{a}{b(a-b)}} \frac{a^a}{b^b (a-b)^{a-b}}$

## Corollary: Normal from Binomial

Let $0 < p < 1$ and $p+q=1$. Let $F_n(x)=\binom{n}{x}p^x q^{n-x}$ $f_n(x)=\sqrt{npq}F_n(np+x\sqrt{npq})$ $\phi_n(x)=\ln(f_n(x)) \\ \\ \phi_n(x)=\ln(n!)-\ln((np+x\sqrt{npq})!)-\ln((nq-x\sqrt{npq})!)+(np+x\sqrt{npq})\ln(p)+(nq-x\sqrt{npq})\ln(q)$ Using Stirling's Approximation and some algebra $\phi_n(x) = -\tfrac{1}{2}\ln(2\pi)-\left (\tfrac{1}{2}+ np+x\sqrt{npq} \right )\ln\left ( 1+x\sqrt{\frac{q}{np}} \right)-\left (\tfrac{1}{2}+ nq-x\sqrt{npq} \right )\ln\left ( 1-x\sqrt{\frac{p}{nq}} \right)+O(\tfrac{1}{n})$ Using the series expansion $\ln(1+x)=x-\tfrac{1}{2}x^2+O(x^3)$ $\phi_n(x) = -\tfrac{1}{2}\ln(2\pi)-\tfrac{1}{2}x^2+O(\tfrac{1}{\sqrt{n}})$ Thus, as $n$ goes to infinity $\phi_\infty(x) = -\tfrac{1}{2}\ln(2\pi)-\tfrac{1}{2}x^2$ $f_\infty(x) = \frac{e^{-x^2/2}}{\sqrt{2\pi}}$ Thus, in the limit, scaling for the changing means and variances, the binomial distribution tends to the normal distribution. Moreover, since the binomial distribution is normalized, we find that $\int_{-\infty}^{\infty}\frac{e^{-x^2/2}}{\sqrt{2\pi}}dx=1$