Find the limit using Stirling's Formula.

Question

The problem:

Let $a_{n} = 1 \cdot 3 \cdot 5\cdot . . . \cdot 2n-1 = \prod^{n}_{k=1} \left(2k-1\right)$.

Then, use Stirling's formula to find $\lim_{n\to\infty} \frac{a_{n}}{\left(\frac{n}{e}\right)^{n} 4^{n} \sqrt{2}}$.

My work so far:

I know that $\prod^{n}_{k=1} \left(2k-1\right)$ can be expressed as $\frac{\left(2n\right)!}{2^{n}n!}$ and I know that Stirling's formula is $n! \approx \left(\frac{n}{e}\right)^{n}\sqrt{2\pi n}$.

Now this is where I am stuck.

I have $\lim_{n\to\infty} \frac{\frac{\left(2n\right)!}{2^{n}n!}}{\left(\frac{n}{e}\right)^{n} 4^{n} \sqrt{2}}$.

Should I replace each $n!$ with $\left(\frac{n}{e}\right)^{n}\sqrt{2\pi n}$, and then try and find the limit, or is there a more efficient way to solve this problem?

Answer 1

Yes, you should just plug the formula in for each factorial. Make sure you get the formula for $(2n)!$ right. Then it is a few lines of algebra. All the $n$s should disappear.

Answer 2

$$(2n)! \approx \left(\frac{2n}{\mathrm{e}}\right)^{2n}\sqrt{2\pi (2n)}$$ $$ 2^n n! \approx 2^n \left(\frac{n}{\mathrm{e}}\right)^n \sqrt{2 \pi n}$$ So $$ \frac{a_n}{\left(\frac{n}{\mathrm{e}} \right)^n 4^n \sqrt{2}} = \frac{\frac{(2n)!}{2^n n!}}{\left(\frac{n}{\mathrm{e}} \right)^n 4^n \sqrt{2}} \approx \frac{ \left( \frac{\left(\frac{2n}{\mathrm{e}}\right)^{2n} \sqrt{2\pi (2n)}}{2^n \left(\frac{n}{\mathrm{e}}\right)^n \sqrt{2 \pi n}}\right) }{\left(\frac{n}{\mathrm{e}} \right)^n 2^{2n} \sqrt{2}}. $$ Simplify and discover that only one use of $n$ remains.