Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I was reading this paper and at the top of page 9 it says that as $n\to\infty$, $$\left(1+\frac{1}{n}\right)^{n+1/2}e^{-1}\left(1+\frac{a_1}{(n+1)}+\frac{a_2}{(n+1)^2}+\cdots \right)=1+\frac{a_1}{n}+\frac {a_2-a_1+1/12}{n^2}+\frac{(13/12) a_1-2a_2+a_3-1/12}{n^3}...$$ I just do not understand where the $1/12,~13/12,~\text{etc.}$ come from, so can anybody enlighten me?

And something else:

If I have shown the exact relation $n!= \sqrt{n}(n/e)^n e^{1-E(n)}$ where $$E(n)=\sum\limits_{k=1}^{n-1} \biggl[\left(\frac {2k+1}{2}\right)\ln \left(\frac{k+1}{k}\right)-1\biggl]$$ that after some working gets to $$\sum\limits_{k=1}^{n-1} \biggl[\left(\frac{1}{3(2k+1)^2}+\frac{1}{5(2k+1)^4}+ \cdots\right)\biggl]<\frac{n-1}{12n},$$ can I use this in any way to derive Stirling's series (the series, not the first term)?

I know that I can derive the series using the Euler-Maclaurin formula but I want this for an essay and I am well off the word limit to introduce a new result.

Thank you.

share|cite|improve this question
A detailed derivation of Stirling's formula can be found in "Concrete Mathematic" by Graham, Knuth and Patashnik (in fact, the derivation is done bit-by-bit all over the book). Does this help? – Johannes Kloos Jan 22 '13 at 21:13
How is the first question related to the second? – Rahul Jan 29 '13 at 5:40
up vote 3 down vote accepted

Stirling formula can be derived using steepest descent.

\[ n! = \int_0^\infty t^n e^{-t} \, dt = \int_0^\infty e^{-t + n \ln t} \, dt\]

The exponent $f(t) = -t + n \ln t$ has critical point at $t = n$ since $f'(t) = -1 + n/t$

\[ f(t) = -n + n \ln n - \frac{1}{2n} (t-n)^2 + O(t-n)^3 \]

This second order expansion means we can approximate the factorial function using a Gaussian integral:

\[ n! \simeq \int_0^\infty e^{-n + n \ln n - \frac{1}{2n} (t-n)^2 } \, dt = e^{-n + n \ln n} \sqrt{2\pi n}\]

If you're more careful, you can get all the terms this way. Nobody really has a good explanation for it, though.

In general, if $x_0$ is a critical point of $f$ then Laplace's method gives

\[ \int_a^b e^{M f(x)} \, dx \approx \sqrt{\frac{ 2\pi }{ M | f''(x_0)|}}e^{M f(x_0)} \]

share|cite|improve this answer

The Euler constant $e$ has a standard representation as the limit, as $n\rightarrow \infty$, of $(1+1/n)^n$. For finite values of $n$, the product in the expression you look at has the expansion $e^{-1}(1+1/n)^n=1-\frac{1}{2n}+\frac{11}{24n^2}\ldots$. The actual factor we have is $e^{-1}(1+1/n)^n(1+1/n)^{1/2}$. Using the series for $(1+x)^{1/2}=1+(1/2)x+\ldots$, (with $x=1/n$) multiplying the series, and collecting terms, you get:$e^{-1}(1+1/n)^n=1+\frac{1}{12n^2}-\frac{1}{12 n^3}\ldots$. This is easy to check in Mathematica but even by hand is not that bad once we use the basic representation of $e$. This calculation gives the terms not involving $a_i$.

As for the $13/12$ and such terms, they come from expanding $\frac{1}{n+1}=\frac{1}{n}\frac{n}{n+1}=\frac{1}{n}\frac{1}{1+(1/n)}=\frac{1}{n}(1-\frac{1}{n}+\frac{1}{n^2}+\ldots$), again, collecting terms and such.

This is an answer to the first part of the question.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.