Stirling's Formula
In permutations, we showed that the number of permutations of \(n\) distinct objects is given by the factorial function \(n!\) How quickly does the factorial function \(n!\) grow as a function of \(n?\) This behavior is captured in the approximation known as Stirling's formula \((\)also known as Stirling's approximation\()\).
Stirling's Formula
The factorial function \(n!\) is approximated by
\[ n! \sim \sqrt{2 \pi n}\left(\frac{n}{e}\right)^n.\]
Furthermore, for any positive integer \(n\), we have the bounds
\[ \sqrt{2\pi}\ n^{n+{\small\frac12}}e^{-n} \le n! \le e\ n^{n+{\small\frac12}}e^{-n}. \]
What is the value of the following limit?
\[ \lim_{n\to\infty} \dfrac{ \sqrt[n]{1 \times 2 \times \cdots \times n} }n \]
Central Binomial Coefficient
The theorem defined in binomial coefficient as \( { 2n \choose n } = \frac { (2n)!} {n!^2} \) for \(n \geq 0 \) and it approaches \( \frac {4^n}{\sqrt{\pi n}} \) asymptotically.
Given Stirling's formula, we have
\[\begin{align} n! &\approx \sqrt{2 \pi n} \left ( \frac {n}{e} \right )^n \\ (2n)! &\approx \sqrt{4 \pi n} \left ( \frac {2n}{e} \right )^{2n}. \end{align} \]
Taking the ratio of the second approximation to the square of the first approximation,
\[ \frac {(2n)!}{n!^2} \implies \sqrt{ \frac {4\pi n}{ (2 \pi n)^2 } } \cdot \frac {(2n)^{2n}}{n^{2n}} \cdot \frac {e^{2n}}{e^{2n}} = \frac {1}{\sqrt{\pi n}} \cdot 2^{2n} = \frac {4^n}{\sqrt{\pi n}}.\ _\square \]
Evaluate
\[ \left \lfloor \frac{2}{1} \times \frac{4}{3} \times \frac{ 6}{5} \times \cdots \times \frac{ 98}{97} \times \frac{ 100}{99} \right \rfloor .\]
Catalan Number
Catalan number occurs many times in various counting problems; it is defined to be \( \large C_n = \frac {1}{n+1} { 2n \choose n } \). For more information, go to Catalan Numbers.
Find the limit
\[\large \displaystyle\lim\limits_{x\rightarrow \infty} \frac{\ln\frac{x^x}{x!}}{x}.\]
Note: For great honor, solve it without WolframAlpha.