# Find the limit using Stirling's Formula.

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?

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.
• Thanks for the help! Just to remove any ambiguity, when you says "all the $n$s should disappear, do you mean for $\left(2n\right)!$, the numerator, the denominator, or all of the above? Aug 31, 2020 at 23:06
$$(2n)! \approx \left(\frac{2n}{\mathrm{e}}\right)^{2n}\sqrt{2\pi (2n)}$$ $$2^n n! \approx 2^n \left(\frac{n}{\mathrm{e}}\right)^n \sqrt{2 \pi n}$$ 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.
• I ended up with $\frac{1}{2^{n}}$. Does this seem right? Aug 31, 2020 at 23:36
• That seems about right. And you can take the limit as $n \rightarrow \infty$? Aug 31, 2020 at 23:39
• Yes. I got an answer of $0$. Aug 31, 2020 at 23:47