Here’s an approximation for *n*! similar to Stirling’s famous approximation.

I ran into this reading A View from the Top by Alex Iosevich. It is less accurate than Stirling’s formula, but has three advantages.

- It contains the highest order terms of Stirling’s formula.
- It’s easy to derive.
- It’s easy to remember.

One way to teach Stirling’s formula would be to teach this one first as a warm-up.

I’ll show that the approximation above gives a lower bound on *n*!. It gives a good approximation to the extent that the inequalities don’t give up too much.

Exponentiating both sides shows

The derivation only uses techniques commonly taught in second-semester calculus classes: using integrals to estimate sums and integration by parts.

Deriving Stirling’s approximation

requires more work, but remembering the result is just a matter of adding an extra factor to the simpler approximation.

As I noted up front, the advantage of the approximation discussed here is simplicity, not accuracy. For more sophisticated approximation, see Two useful asymptotic series for how to compute the logarithm of the gamma function.

**Update**:

The first comment asked whether you tweak the derivation to get a simple *upper* bound on factorial. Yes you can.

This leads to the upper bound

Wasteful compared to Stirling, but still simple to derive and remember.

**Related posts**:

David Mackay has an unconventional derivation of Stirling approximation. He approximate the Poisson distribution of parameter Lambda by a Gaussian of mean and variance Lambda (good approximation for big values of Lambda). The approximation follows from that.

See page 2 of his book “Information theory …”, or page 14 of the PDF, available here:

http://www.inference.phy.cam.ac.uk/mackay/itila/book.html

for the details.

Nice. Can you tweak the derivation to get a simple upper bound?

Yes. The original derivation uses an integral to under-estimate the sum. You can shift the sum over by one and use an integral to over-estimate the sum.

Thanks for the suggestion. I’ve updated the post with an upper bound.

It would seem more useful (and just as simple) if you don’t drop the (+1) in the third line. You’d get n! gte e(n/e)^n instead of n! gt (n/e)^n

Nice again! Thanks.

You can get the square root of n factor just by using the trapezoidal rule in your derivation, instead of a Riemann sum.