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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?

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2 Answers 2

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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.

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  • $\begingroup$ 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? $\endgroup$ Aug 31, 2020 at 23:06
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    $\begingroup$ All of the above. If they didn't you wouldn't have a nonzero limit at infinity. The expression in the problem was chosen to make that so. $\endgroup$ Aug 31, 2020 at 23:19
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$$(2n)! \approx \left(\frac{2n}{\mathrm{e}}\right)^{2n}\sqrt{2\pi (2n)}$$ $$ 2^n n! \approx 2^n \left(\frac{n}{\mathrm{e}}\right)^n \sqrt{2 \pi n}$$ 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.

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  • $\begingroup$ I ended up with $\frac{1}{2^{n}}$. Does this seem right? $\endgroup$ Aug 31, 2020 at 23:36
  • $\begingroup$ That seems about right. And you can take the limit as $n \rightarrow \infty$? $\endgroup$ Aug 31, 2020 at 23:39
  • $\begingroup$ Yes. I got an answer of $0$. $\endgroup$ Aug 31, 2020 at 23:47
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    $\begingroup$ @MiltonP. : Woot! $\endgroup$ Aug 31, 2020 at 23:50
  • $\begingroup$ Thank you for the help! $\endgroup$ Sep 1, 2020 at 14:12

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