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3.1 Sums & Products

Applying Stirling's Formula


The quantity

\Large \frac{(2n)!}{2^{2n}(n!)^2} \qquad\qquad

will come up later in the course (it is the probability that in 2^{2n} flips of a fair coin, exactly n will be Heads). Which of the following simplified formulas is asymptotically equal to the above formula?

(Hint: to find the final formula, write out a short sequence of successively simpler formulas.)

Exercise 1
\begin{array}{rlc} \frac{(2n)!}{ 2^{2n}(n!)^2} & \sim \frac{(2n/e)^{2n}\sqrt{2 \pi 2n}} {2^{2n}\left[(n/e)^n\sqrt{2 \pi n}\right]^2} & \text{(by Stirling's Formula)}\\ & = \frac{2^{2n}(n/e)^{2n}\sqrt{2 \pi 2n}}{2^{2n} (n/e)^{2n} \left[\sqrt{2 \pi n}\right]^2}\\ & = \frac{\sqrt{2 \pi 2n}} {\left[\sqrt(2 \pi n)\right]^2}\\ & = \frac{\sqrt{2} \sqrt{2 \pi n}}{\left[\sqrt{2 \pi n}\right]^2}\\ & = \frac{\sqrt{2}}{\sqrt{2 \pi n}} = \frac{1}{\sqrt{\pi n}} \end{array}