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i cant see why we have :

  • $$\frac{(2n)!}{4^n n!^2} = \frac{(2n-1)!!}{(2n)!!}$$

  • $$\dfrac{(2n-1)!!}{(2n)!!} =\prod_{k=1}^{n}\left(1-\dfrac{1}{2k}\right),$$

Even i see the notion of Double factorial

this question is related to that one : Behaviour of the sequence $u_n = \frac{\sqrt{n}}{4^n}\binom{2n}{n}$

  • For $\frac{(2n)!}{4^n n!^2} = \frac{(2n-1)!!}{(2n)!!}$

$\dfrac{(2n)!}{4^n n!^2}=\dfrac{(2n)!}{2^{2n} n!^2}=\dfrac{(2n)\times (2n-1)!}{2^{2n} (n\times (n-2)!)^2}$

  • For $\dfrac{(2n-1)!!}{(2n)!!} =\prod_{k=1}^{n}\left(1-\dfrac{1}{2k}\right),$

note that $n!! = \prod_{i=0}^k (n-2i) = n (n-2) (n-4) \cdots$

$\dfrac{(2n-1)!!}{(2n)!!}=\dfrac{\prod_{i=0}^k (2n-1-2i)}{\prod_{i=0}^k (2n-2i)}$

Community
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Educ
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2 Answers2

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Note first that $(2n)!=(2n)!!(2n-1)!!$, because $(2n)!!$ gives you the even factors in $(2n)!$, and $(2n-1)!!$ gives you the odd factors. Now

$$\begin{align*} (2n)!!&=(2n)(2n-2)(2n-4)\ldots(4)(2)\\ &=\big(2n\big)\big(2(n-1)\big)\big(2(n-2)\big)\ldots\big(2(2)\big)\big(2(1)\big)\\ &=2^nn!\;, \end{align*}$$

so

$$\frac{(2n)!}{4^nn!^2}=\frac{(2n)!!(2n-1)!!}{2^{2n}n!^2}=\frac{2^nn!(2n-1)!!}{2^{2n}n!^2}=\frac{(2n-1)!!}{2^nn!}=\frac{(2n-1)!!}{(2n)!!}\;.$$

Now

$$\begin{align*} \frac{(2n-1)!!}{(2n)!!}&=\frac{2n-1}{2n}\cdot\frac{2n-3}{2n-2}\cdot\frac{2n-5}{2n-4}\cdot\ldots\cdot\frac34\cdot\frac12\\\\ &=\left(1-\frac1{2n}\right)\left(1-\frac1{2n-2}\right)\left(1-\frac1{2n-4}\right)\ldots\left(1-\frac14\right)\left(1-\frac12\right)\\\\ &=\prod_{k=1}^n\left(1-\frac1{2k}\right)\;. \end{align*}$$

Brian M. Scott
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  • @BrianMScott Out of curiosity, is there a reason why you use \ldots rather than the conventional \cdots during multiplication constructions? I find that \cdots yields a more typographically pleasing output, but maybe you prefer the other construction. – Daniel W. Farlow Jan 07 '15 at 16:37
  • @induktio: My æsthetics are the opposite of yours here (and it seems to me that both versions are standard/conventional). I dislike having the ellipsis run together with the operation. – Brian M. Scott Jan 07 '15 at 16:40
  • @BrianMScott Interesting! I just came across the following in the TeXbook in the section about the fine points of mathematical typesetting: "In general, it is best to use \cdots between $+$ and $-$ and $\times$ signs, and also between $=$ signs or $\leq$ signs or $\subset$ signs or other similar relations. Low dots are used between commas, and when things are juxtaposed with no signs between them at all" (pg. 172). He gives this example: $(1-x)(1-x^2)\ldots (1-x^n)$. Neat. I guess everything I have read before has been typeset incorrectly. – Daniel W. Farlow Jan 07 '15 at 16:48
  • @induktio: To be honest, I prefer (and use) the low dots in almost all settings. This may be because I’m older, and my æsthetics derive in large part from printed material that’s older than computer typesetting. – Brian M. Scott Jan 07 '15 at 16:52
  • That's what we call detailed answer – Educ Jan 07 '15 at 17:02
  • could you explain to me why Jason wrote $2n!!=\prod_{k=1}^n(2k)$ i cant see from wiki $n!!=\prod_{i=0}^k (n-2i)$ so $2n!!=\prod_{k=1}^k(2n-2i)$ – Educ Jan 07 '15 at 17:27
  • @Educ: That should be $(2n)!!=\prod_{k=1}^n(2k)$: the parentheses are crucial. $\prod_{k=1}^n(2k)$ and $\prod_{i=0}^{2n}(2n-2i)$ are the same products in reverse order: the first is $(2)(4)\ldots(2n-2)(2n)$, and the second is $(2n)(2n-2)\ldots(2)$. – Brian M. Scott Jan 07 '15 at 17:36
  • in general we can right $n!!=\prod_{k=1}^n(2k)=\prod_{k=1}^n(n-k)$ – Educ Jan 07 '15 at 17:39
  • @Educ: No: $\prod_{k=1}^n(2k)$ is $(2n)!!$, not $n!!$, and $\prod_{k=1}^n(n-k)=0$, since the last factor is $n-n=0$. What is true is that $$n!!=\prod_{k=0}^{\lfloor n/2\rfloor}(n-2k);.$$ – Brian M. Scott Jan 07 '15 at 17:42
  • and this is gonna be true only for $2n!!$ and $(2n-1)!!$ – Educ Jan 07 '15 at 17:44
  • @Educ: I’m not sure what your this refers to. The displayed formula in my last comment is valid for all positive integers $n$. – Brian M. Scott Jan 07 '15 at 17:47
  • i meant $(2n-1)!!=\prod_{k=1}^n(2k-1)$ and $(2n)!!=\prod_{k=1}^n(2k)$ – Educ Jan 07 '15 at 17:55
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    @Educ: Ah, okay; then you’re absolutely right. – Brian M. Scott Jan 07 '15 at 17:56
  • just to be sure have we : $(2n)!!=\prod_{k=1}^n(2k)=\prod_{i=0}^{2n}(2n-2i)$ and $(2n-1)!!=\prod_{k=1}^n(2k-1)=\prod_{i=0}^{2n-1}(2n-1-2i)$ – Educ Jan 07 '15 at 18:03
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    @Educ: Not quite: the final product in the first case should be $\prod_{i=0}^{n-1}(2n-2i)$, so that the last factor is $2$, and the final product in the second case should be $\prod_{i=0}^{n-1}(2n-1-2i)$, so that the last factor is $1$. – Brian M. Scott Jan 07 '15 at 18:08
  • there is typo at one of ur comments : Educ: That should be $(2n)!!=\prod_{k=1}^n(2k)$ : the parentheses are crucial.$\prod_{k=1}^n(2k)$ and $\prod_{i=0}^{2n}(2n-2i)$ here in that $\prod_{i=0}^{2n}(2n-2i)$ as you said should be $\prod_{i=0}^{n-1}(2n-2i)$ – Educ Jan 07 '15 at 18:21
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    @Educ: Good catch. Yes, that was a careless typo on my part. – Brian M. Scott Jan 07 '15 at 18:22
  • i think ur definition of double factorial is not quite correct $n!!=\prod_{k=0}^{\lfloor n/2\rfloor}(n-2k);.$ the right one is $n!!=\prod_{k=0}^{k=\lceil n/2 \rceil - 1.}(n-2k);.$ with ceil function not with floor – Educ Jan 07 '15 at 18:36
  • @Educ: Oops. Yes, you’re right. What I wrote works fine for odd $n$ but gives $0$ when $n$ is even. And at this point I think that you can safely say that you’re on top of the subject! :-) – Brian M. Scott Jan 07 '15 at 18:41
  • @BrianM.Scott Thanks you for ur time and ur detailed explantion :) – Educ Jan 07 '15 at 18:43
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    @Educ: You’re welcome. Sorry about the careless bits! – Brian M. Scott Jan 07 '15 at 18:45
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$\dfrac{(2n)!}{4^nn!^2}=\dfrac{(2n)!!(2n-1)!!}{(2^nn!)^2}=\dfrac{(2n)!!(2n-1)!!}{((2n)!!)^2}=\dfrac{(2n-1)!!}{(2n)!!}$

$\prod_{k=1}^n(1-\frac{1}{2k})=\prod_{k=1}^n\frac{2k-1}{2k}=\dfrac{\prod_{k=1}^n(2k-1)}{\prod_{k=1}^n(2k)}=\frac{(2n-1)!!}{(2n)!!}$

MMM
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  • it's not detailed – Educ Jan 07 '15 at 16:29
  • from wiki $n!!=\prod_{i=0}^k (n-2i)$ so $2n!!=\prod_{k=1}^k(2n-2i)$ why we hace $2n!!=\prod_{k=1}^n(2k)$ – Educ Jan 07 '15 at 17:23
  • @Educ well, the product of $2n-2i$ where $i$ from $1$ to $n$ is equal to the product of $2k$ where $k$ from $1$ to $n$ (or $n$ to $1$), as long as you let $k=n-i$ – MMM Jan 19 '15 at 16:02