(Sheet 7) Pudding - Stirling's formula

Below are two questions from an ancient (1970) first-year exam paper, which together lead you through a proof of Stirling's approximation for \( n! \).

A scan of the original paper is available (Stirling questions), in case you would like to see how the original questions appeared.  This also provides a printable version of the questions, if you prefer.  (Note that only questions 2 and 3 from this page are relevant!)

2. Show that

\[ \sum_{r=1}^{\infty} \left \{ \int_r^{r+1} \log x \ dx - \frac{1}{2} [\log r + \log(r+1)] \right\} \]

is a convergent series.

Hence, or otherwise, show the existence of

\[ \lim_{n \to \infty} \frac{n!}{n^{n+\frac{1}{2}} e^{-n}}. \]

3. Let \( \displaystyle I_p = \int_0^{\frac{1}{2}\pi} (\sin x)^p dx \).

1. Show that \(I_{2n-1} > I_{2n} > I_{2n+1} \) for \( n \geq 1 \).
2. Show, by integration by parts, that \( \displaystyle I_p = \frac{p-1}{p} I_{p-2} \) for \( p \geq 2\).  Deduce that
   \[ \begin{eqnarray*} I_{2n} & = \frac{2n-1}{2n} \cdot \frac{2n-3}{2n-2} \dotsm \frac{1}{2} \frac{\pi}{2} \\ I_{2n+1} & = \frac{2n}{2n+1} \cdot \frac{2n-2}{2n-1} \dotsm \frac{2}{3}. \end{eqnarray*} \]
3. From (i) and (ii) deduce that \( \displaystyle \lim_{n \to \infty} \frac{I_{2n}}{I_{2n+1}} = 1 \) and so prove that
   \[ \frac{\pi}{2} = \lim_{n \to \infty} \frac{2^{4n}(n!)^4}{[(2n)!]^2(2n+1)}. \]
4. Assume that the following limit exists and use (iii) to prove that
   \[ \lim_{n \to \infty} \frac{n!}{n^{n+\frac{1}{2}} e^{-n} } = \sqrt{2 \pi}. \]