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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?

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

1 Answer 1

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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

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  • $\begingroup$ Wow. Thank you. $\endgroup$ Feb 7, 2022 at 18:21
  • $\begingroup$ Ok wow, this looked like a cute problem from first sight tho. $\endgroup$ Feb 7, 2022 at 18:39
  • $\begingroup$ Shouldn't we have $$\binom{2n}{n}\sim\frac{4^n}{\sqrt{n\pi}}?$$ $\endgroup$ Feb 7, 2022 at 19:32
  • $\begingroup$ Yes, you right. I fixed it. Thanks @Aaron Hendrickson. $\endgroup$
    – boaz
    Feb 8, 2022 at 20:35

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