# Find the limit of $u_n = \frac{1\times3\times5\times7\times..........\times(2n-1)}{2\times4\times6\times8\times........\times(2n)}$

To prove that $$u_n = \frac{1\times3\times5\times7\times..........\times(2n-1)}{2\times4\times6\times8\times........\times(2n)}$$ converges, we can notice that $$1/2$$ is the smallest term in the product and $$(\frac{2n-1}{2n})$$ is the greatest.

Hence : $$(1/2)^{2n-1} \leq u_n \leq (\frac{2n-1}{2n})^{2n-1}$$. By squeeze theorem we see that $$\lim u_n = 0$$.

Is this solution correct?

• $lim_{n\to\infty}(\frac{2n-1}{2n})^{2n-1}=\frac{1}{e}$ and not zero, so your solution is incorrect.
– boaz
Feb 7, 2022 at 18:13
• @boaz How can I do it then? Feb 7, 2022 at 18:16
• $u_n=\frac{1\times 2\times\dots\times(2n-1)\times(2n)}{(2\times 4\times\dots\times (2n))^2}=\frac{(2n)!}{2^{2n}\cdot(n!)^2}$. Then use Stirling's approximation of the factorial expresions. Feb 7, 2022 at 18:22

Hint (corrected): Note that $$\frac{(1)(3)(5)\ldots(2n-1)} {(2)(4)(6)\ldots(2n)}= \frac{(1)(2)(3)(4)(5)\ldots(2n-1)(2n)} {[(2)(4)(6)\ldots(2n)]^2}= \frac{(2n)!}{2^{2n}(n!)^2} =\frac{\binom{2n}{n}}{4^n}$$ Now use the estimate $$\binom{2n}{n}\sim\frac{4^n}{\sqrt{n\pi}}$$ see https://en.wikipedia.org/wiki/Central_binomial_coefficient
• Shouldn't we have $$\binom{2n}{n}\sim\frac{4^n}{\sqrt{n\pi}}?$$ Feb 7, 2022 at 19:32