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Ok, so I want to show that

$$\lim_{n\to\infty}\frac{1\cdot 3\cdot 5\cdots (2n-1)}{(2n)^n} = 0.$$

Here is what I have tried so far:

\begin{align} \notag \lim_{n\to\infty}\frac{1\cdot 3\cdot 5\cdot \cdots \cdot (2n-1)}{(2n)^n} &= \lim_{n\to\infty}\frac{\prod_{j=1}^{n}(2j-1)}{(2n)^n}\\ \notag &= \lim_{n\to\infty}\frac{\prod_{j=1}^{n}(2j-1)}{(2n)^n}\cdot\frac{\prod_{j=1}^{n}(2j)}{\prod_{j=1}^{n}(2j)}\\ \notag &= \lim_{n\to\infty}\frac{\prod_{j=1}^{2n}j}{(2n)^n\prod_{j=1}^{n}(2j)}\\ \notag &= \lim_{n\to\infty}\frac{(2n)!}{(2n)^n\prod_{j=1}^{n}(2j)}\\ \notag &= \lim_{n\to\infty}\frac{(2n)!}{(2n)^n\cdot 2^n\cdot n!}\\ \notag &= \lim_{n\to\infty}\frac{(n+1)\cdot\cdots\cdot (2n)}{(4n)^{n}} \end{align}

From here you can tell that it vanishes, but is this enough to show that the limit is 0? Is there any way to break it down further? Additionally, are there any handy product identities to know? Basically anything analogous to identities like the following:

$$\sum_{i=1}^n = \frac{n(n+1)}{2}$$

Thanks

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HINT:

$$\frac{1\cdot 3\cdot 5\cdots (2n-1)}{(2n)^n}=\prod_{1\le r\le n}\frac{2r-1}{2n}<\frac1{2n} $$

as each term $\le1$ and $2n>2r-1$ for finite $r$

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    $\begingroup$ Excellent, I'll have to remember to use inequalities more often, thanks! $\endgroup$ – chs21259 Dec 28 '13 at 18:43
  • $\begingroup$ @chs21259, welcome. But, not sure why you have to remember? Just observe the multipliers $\endgroup$ – lab bhattacharjee Dec 28 '13 at 18:49
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What about this? $$ \lim_{n \to \infty} \frac{1 \cdot 3 \cdot \dots \cdot 2n-1}{(2n)(2n) \cdots (2n)} \le \lim_{n \to \infty} \frac 1{2n} \cdot 1 \cdot \dots \cdot 1 = 0 $$

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  • $\begingroup$ Looks like I used a shotgun to kill a fly. This is exactly what I was looking for, thanks! $\endgroup$ – chs21259 Dec 28 '13 at 18:43
  • $\begingroup$ To be honest you just "multiplied by $1$", you didn't do much. I think you just grabbed the shotgun. $\endgroup$ – Patrick Da Silva Dec 28 '13 at 18:52
  • $\begingroup$ All I'm saying is that your approach seemed much simpler than mine. $\endgroup$ – chs21259 Dec 28 '13 at 19:00
  • $\begingroup$ @chs21259 : Speaking of useful identities though, I'm a good fan of $1 \cdot 3 \cdot \cdots \cdot 2n-1 = (2n)! / (2^n n!)$. Then you can use Stirling's approximations for the factorials. $\endgroup$ – Patrick Da Silva Dec 28 '13 at 19:04
  • $\begingroup$ Interesting! That will come in handy, thanks! $\endgroup$ – chs21259 Dec 28 '13 at 19:10
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Another approach:

$$a_n:=\frac{1\cdot 3\cdot\ldots\cdot(2n-1)}{2^nn^n}=\frac{(2n)!}{2^{2n}n!n^n}$$

so

$$\frac{a_{n+1}}{a_n}=\frac{(2n+2)!}{2^{2n+2}(n+1)!(n+1)^{n+1}}\frac{2^nn!n^n}{(2n)!}=\frac12\frac{2n+1}{n+1}\frac1{\left(1+\frac1n\right)^n}\xrightarrow[n\to\infty]{}\frac1e<1\implies$$

$$\implies\sum_{n=1}a_n\;\;\text{converges}\;\implies a_n\xrightarrow[n\to\infty]{}0$$

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  • $\begingroup$ So, just to be clear, you used, Ratio Test -> Divergence Theorem? That's pretty clever! $\endgroup$ – chs21259 Dec 28 '13 at 19:15
  • $\begingroup$ Yes...though in this case it was Ratio test --> convergence theorem. $\endgroup$ – DonAntonio Dec 28 '13 at 19:17
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One more way: rewrite numerator as $(2n-1)!!=\frac{(2n)!}{2^{n}n!}$ and denominator using Stirling's formula as $(2n)^n=(2n)^{\frac{2n}{2}}=e^{2n} (2n)! 2\sqrt{\pi n}(1+o(1))$. Hence your expression becomes $$ S=\frac{(2n)!}{2^n n! e^{2n} (2n)! 2\sqrt{\pi n}(1+o(1))}=\frac{1}{2^n n! e^{2n} (2n)! 2\sqrt{\pi n}(1+o(1))} $$ which happily tends to $0$.

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  • $\begingroup$ Ah, Patrick just suggested this, nice to see it worked out. Thanks! $\endgroup$ – chs21259 Dec 28 '13 at 19:11
  • $\begingroup$ You are welcome $\endgroup$ – Alex Dec 29 '13 at 3:05
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My hint:

$$\frac{(2n-1)!!}{(2n)^n}=\frac{(2n)!}{n!(4n)^n}\sim \frac{(\frac{2n}{e})^{2n}\sqrt{4\pi n}}{(\frac{n}{e})^{n}\sqrt{2\pi n}(4n)^n}=\frac{\sqrt{2}(2n)^n}{(2ne)^n}=\sqrt{2}\: \left(\frac{1}{e}\right)^n\to 0$$

Hence: $$\lim_{n\to \infty}\frac{(2n-1)!!}{(2n)^n}=0$$

I use Stirling's approximation

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