Stirling's Theorem - Speed of Convergence and Error Estimates

Speed of Convergence and Error Estimates

More precisely,

with

Stirling's formula is in fact the first approximation to the following series (now called the Stirling series):

\begin{align} n! &\sim \sqrt{2\pi n}\left(\frac{n}{e}\right)^n \left(1 +{1\over12n}+{1\over288n^2} - {139\over51840n^3} -{571\over2488320n^4}+ \cdots \right) \\ &= \sqrt{2\pi n}\left(\frac{n}{e}\right)^n \left(1+\frac{1}{(2^1)(6n)^1}+{1\over(2^3)(6n)^2}-{139\over(2^3)(2\cdot3\cdot5)(6n)^3} \right. \\ &\qquad\left. -{571\over(2^6)(2\cdot3\cdot5)(6n)^4} + \cdots \right). \end{align}

Explicit formula for the coefficients in this series was given by G. Nemes. The first graph in this section shows the relative error vs. n, for 1 through all 5 terms listed above.

As n → ∞, the error in the truncated series is asymptotically equal to the first omitted term. This is an example of an asymptotic expansion. It is not a convergent series; for any particular value of n there are only so many terms of the series that improve accuracy, after which point accuracy actually gets worse. This is shown in the next graph, which shows the relative error versus the number of terms in the series, for larger numbers of terms. More precisely, let S(n, t) be the Stirling series to t terms evaluated at n. The graphs show

which, when small, is essentially the relative error.

Writing Stirling's series in the form:

\begin{align} \ln(n!) &\sim n\ln(n) - n + \tfrac{1}{2}\ln(2\pi n) +{1\over12n} -{1\over360n^3} +{1\over1260n^5} -{1\over 1680n^7} +\cdots \\ &= n\ln(n)-n+\tfrac{1}{2}\ln(2\pi n)+{1\over(2^2\cdot3^1)n}-{1\over(2^3\cdot3^2\cdot5^1)n^3}+{1\over(2^2\cdot3^2\cdot5^1\cdot7^1)n^5}\\ &\qquad - \frac{1}{(2^4 \cdot3^1 \cdot5^1\cdot7^1)n^7} +\cdots. \end{align}

it is known that the error in truncating the series is always of the same sign and at most the same magnitude as the first omitted term.

Read more about this topic:  Stirling's Theorem

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