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$$ \lim_ {n\to\infty} \dfrac {\left [\left (2n-1\right)!! \right] ^ {1/ {2n}}} {\left [\displaystyle\prod_ {k=1} ^ {n} (2k-1)!! \right] ^ {1/ {n^2}}}$$

Daniele Tampieri
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SHZ
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  • Hi and welcome to the Math.SE: what did you tried? In order for the question to be noted and answered, it is a better practice to provide some context. For example, the strategies of solution you tried and failed, why is this problem interesting for you, and so on. – Daniele Tampieri Aug 22 '19 at 08:57
  • For future reference, the value of this limit is $$e^{1/4}=\sqrt{\sqrt{e}}$$. – Peter Foreman Aug 22 '19 at 09:33
  • Yes, the answer is $\text{e}^{1/4}$ – SHZ Aug 22 '19 at 09:51

2 Answers2

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Using Stirling formula $\sqrt{2\pi}n^{n+\frac12}e^{-n}\leq n! \leq \sqrt{2\pi}n^{n+\frac12}e^{-n+\frac1{12n}}$ and $(2n-1)!!=\frac{(2n-1)!}{(2n-2)!}$

Zhaohui Du
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    $\frac{(2n-1)!}{(2n-2)!}=2n-1$. – Feng Aug 22 '19 at 09:08
  • Yes. It should be $(2n-1)!!=\frac{(2n-1)!}{(2n-2)!!}=\frac{(2n-1)!}{2^{n-1}(n-1)!}$ – Zhaohui Du Aug 22 '19 at 09:12
  • $$ \text{e}^{-{1}/{2}}\underset{n\rightarrow \infty}{\lim}\dfrac{\left[ \dfrac{\left( 2n-1 \right) ^{4n-1}}{\left( n-1 \right) ^{2n-1}} \right] ^{{1}/{4n}}}{\left[ \displaystyle\prod_{k=1}^n{\dfrac{\left( 2k-1 \right) ^{4k-1}}{\left( k-1 \right) ^{2k-1}}} \right] ^{{1}/{2n^2}}} $$I've used $n!=\sqrt{2n\pi}\left( \dfrac{n}{\text{e}} \right) ^n\text{e}^{{\theta}/{12n}} $. how to go on – SHZ Aug 22 '19 at 10:15
  • Using $(2n-1)!!=\frac{(2n-1)!}{2^{n-1}(n-1)!}=\frac{(2n)!}{2^n n!}$, and Stirling formula, we could transform the limit into $\lim_{n\to\infty}e^{\frac{\log(n)}{2}-\frac{1}{n^2}\sum_{k=1}^nk\log(k)}$ so it is $\lim_{n\to\infty}e^{\frac{-\log(n)}{2n}-\frac{1}{n}\sum_{k=1}^n\frac{k}{n}\log(\frac{k}{n})}=e^{-\int_0^1 x\log(x)dx}=e^{1/4}$ – Zhaohui Du Aug 23 '19 at 00:11
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The numerator is $\int_0^1 x\log(x)dx=-1/4$ so $\lim_{n\to\infty}\frac{1}{n}(\sum_{k=1}^n \frac{2k-1}{2n-1}\log(\frac{2k-1}{2n-1}))=-\frac14$ so $\lim_{n\to\infty}(\prod_{k=1}^n(\frac{2k-1}{2n-1})^{2k-1})^{\frac{1}{n(2n-1)}}=e^{-\frac14}$

Zhaohui Du
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