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