Editing Stirling's approximation (section)

== Derivation ==
Roughly speaking, the simplest version of Stirling's formula can be quickly obtained by approximating the sum
<math display=block>\ln(n!) = \sum_{j=1}^n \ln j</math>
with an [[integral]]:
<math display=block>\sum_{j=1}^n \ln j \approx \int_1^n \ln x \,{\rm d}x = n\ln n - n + 1.</math>

The full formula, together with precise estimates of its error, can be derived as follows. Instead of approximating <math>n!</math>, one considers its [[natural logarithm]], as this is a [[slowly varying function]]:
<math display=block>\ln(n!) = \ln 1 + \ln 2 + \cdots + \ln n.</math>

The right-hand side of this equation minus
<math display=block>\tfrac{1}{2}(\ln 1 + \ln n) = \tfrac{1}{2}\ln n</math>
is the approximation by the [[trapezoid rule]] of the integral
<math display=block>\ln(n!) - \tfrac{1}{2}\ln n \approx \int_1^n \ln x\,{\rm d}x = n \ln n - n + 1,</math>

and the error in this approximation is given by the [[Euler–Maclaurin formula]]:
<math display=block>\begin{align}
\ln(n!) - \tfrac{1}{2}\ln n & = \tfrac{1}{2}\ln 1 + \ln 2 + \ln 3 + \cdots + \ln(n-1) + \tfrac{1}{2}\ln n\\
& = n \ln n - n + 1 + \sum_{k=2}^{m} \frac{(-1)^k B_k}{k(k-1)} \left( \frac{1}{n^{k-1}} - 1 \right) + R_{m,n},
\end{align}</math>

where <math>B_k</math> is a [[Bernoulli number]], and {{math|''R''<sub>''m'',''n''</sub>}} is the remainder term in the Euler–Maclaurin formula. Take limits to find that
<math display=block>\lim_{n \to \infty} \left( \ln(n!) - n \ln n + n - \tfrac{1}{2}\ln n \right) = 1 - \sum_{k=2}^{m} \frac{(-1)^k B_k}{k(k-1)} + \lim_{n \to \infty} R_{m,n}.</math>

Denote this limit as <math>y</math>.  Because the remainder {{math|''R''<sub>''m'',''n''</sub>}} in the Euler–Maclaurin formula satisfies
<math display=block>R_{m,n} = \lim_{n \to \infty} R_{m,n} + O \left( \frac{1}{n^m} \right),</math>

where [[big-O notation]] is used, combining the equations above yields the approximation formula in its logarithmic form:
<math display=block>\ln(n!) = n \ln \left( \frac{n}{e} \right) + \tfrac{1}{2}\ln n + y + \sum_{k=2}^{m} \frac{(-1)^k B_k}{k(k-1)n^{k-1}} + O \left( \frac{1}{n^m} \right).</math>

Taking the exponential of both sides and choosing any positive integer <math>m</math>, one obtains a formula involving an unknown quantity <math>e^y</math>. For {{math|''m'' {{=}} 1}}, the formula is
<math display=block>n! = e^y \sqrt{n} \left( \frac{n}{e} \right)^n \left( 1 + O \left( \frac{1}{n} \right) \right).</math>

The quantity <math>e^y</math> can be found by taking the limit on both sides as <math>n</math> tends to infinity and using [[Wallis product|Wallis' product]], which shows that <math>e^y=\sqrt{2\pi}</math>. Therefore, one obtains Stirling's formula:
<math display=block>n! = \sqrt{2 \pi n} \left( \frac{n}{e} \right)^n \left( 1 + O \left( \frac{1}{n} \right) \right).</math>