A calculus of the absurd
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10.11.2 A lower bound for the factorial function
This is a really cool method (which I discovered originally https://terrytao.wordpress.com/2010/01/02/254a-notes-0a-stirlings-formula/on Terence Tao’s blog) to find a lower bound for the factorial function8080 The factorial function is defined as \(n! = n \cdot (n-1) \cdot ... \cdot 1\). The first step is to consider the Maclaurin series for \(e^x\), i.e.
\(\seteqnumber{0}{10.}{124}\)\begin{equation} e^x = \sum _{n \geqq 0} \frac {x^n}{n!} \end{equation}
If we specify that \(x > 0\), then all the terms in the series are positive, so any individual term of the series will be smaller than the term we have selected and thus we can select the term where \(x=n\) (\(n > 0\)) which satisfies
\(\seteqnumber{0}{10.}{125}\)\begin{equation} e^x \geqq \frac {n^n}{n!} \end{equation}
Therefore,
\(\seteqnumber{0}{10.}{126}\)\begin{equation} n! \geqq \frac {n^n}{e^x} \end{equation}
This is closely related to Stirling’s formula (a closed-form approximation for the factorials which shows up in quite a few places) - the linked blog post gives the full derivation.