How to prove that $\lim_{n \to\infty} \frac{(2n-1)!!}{(2n)!!}=0$

Question

So guys, how can I evaluate and prove that $$\lim_{n \to\infty} \frac{(2n-1)!!}{(2n)!!}=0.$$ Any ideas are welcomed.

$n!!$ is the double factorial, as explained in this wolfram post.

Do you intend to use the double factorial, or simply the factorial of the factorial? I'd assume the double factorial, but I'm just checking... — apnorton, May 09 '14 at 16:34
An involved way that will work: Rewrite everything to pure factorials and use Sterlings approximation. — Winther, May 09 '14 at 16:34
$$\lim_{n\to\infty}\frac{(2n-1)!!}{(2n)!!}=0,\qquad\quad\lim_{n\to\infty}\frac{(2n+1)!!}{(2n)!!}=\infty,\qquad\quad\lim_{n\to\infty}\frac{(2n-1)!!\cdot(2n+1)!!}{(2n)!!\cdot(2n)!!}=\frac2\pi$$ — Lucian, May 09 '14 at 20:50
Take the log. You'll get $\sum_i \log\frac{2i-1}{2i}\approx \sum_i -\frac 1 {2i}=-\infty$, hence the original limit is $0$. — A.S., Dec 01 '15 at 13:31
Speculative: $(2n-1) < 2n$ thus $(2n-1)!! < (2n)!!$ thence the denominator goes to infinity faster than the numerator (they have also the same degree). Thus as $n$ approaches to infinity the denominator "wins", and the limit is zero. — Enrico M., Dec 01 '15 at 13:35
@KimPeekII Yeah, I realise that but it's not really a good proof. — user1640736, Dec 01 '15 at 13:36
I know :) I was just speculating a bit, if you want a rigorous proof, this can only gives you the "idea" of what you should obtain in the end. I'm writing down some idea right now, I'll write an answer later! — Enrico M., Dec 01 '15 at 13:37
@DanielFischer Never heard of it, but would be nice for my own knowledge. — user1640736, Dec 01 '15 at 13:48
Stirling formula applies for the simple factorial as:
$$n! \approx \sqrt{2\pi n}\left(\frac{n}{e}\right)^n$$ — Enrico M., Dec 01 '15 at 13:51
Stirling's formula says $n! \sim \sqrt{2\pi n} \bigl(\frac{n}{e}\bigr)^n$. There are further terms, but those suffice here. Then you can rewrite $$\frac{(2n-1)!!}{(2n)!!} = \frac{1}{2^{2n}}\binom{2n}{n}$$ (or, depending on the used convention for $(2n)!!$, you get some constant factor times that), and with Stirling's formula you get the asymptotic $\frac{(2n-1)!!}{(2n)!!} \sim \frac{1}{\sqrt{\pi n}}$ (or that times a constant). — Daniel Fischer, Dec 01 '15 at 13:54
See also: http://math.stackexchange.com/questions/1554790/evaluate-lim-limits-n-to-infty-frac2n-12n — Martin Sleziak, Dec 01 '15 at 15:12
All: A question merger explains why this question "suddenly" got more answers in Dec'15. — Jyrki Lahtonen, Dec 02 '15 at 07:31
See also: Another evaluating limit question: $\lim\frac{1\cdot3\cdot5\cdot\ldots\cdot(2n-1)}{2\cdot4\cdot6\cdot\ldots\cdot2n}$ — Martin Sleziak, May 13 '18 at 10:20
I'm voting to close this question as off-topic because missing context. — user, Dec 01 '18 at 14:26

score 10 · Answer 1 · edited Apr 13 '17 at 12:21

10

You can notice that $$\frac{(2n-1)!!}{(2n)!!} = \prod_{k=1}^n \frac{2k-1}{2k} = \prod_{k=1}^n \left(1-\frac{1}{2k}\right).$$

So you want in fact calculate the infinite product $$\lim_{n\to\infty} \prod_{k=1}^n \left(1-\frac{1}{2k}\right) = \prod_{k=1}^\infty \left(1-\frac{1}{2k}\right)$$

Now you can use this fact: How to prove $\prod_{i=1}^{\infty} (1-a_n) = 0$ iff $\sum_{i=1}^{\infty} a_n = \infty$?

In fact, for this you do not need the equivalence, only one implication is sufficient: $$\prod_{k=1}^n (1-a_k) \le \frac1{\prod_{k=1}^n (1+a_k)} \le \frac1{1+\sum_{k=1}^n a_k}.$$ (We assume that $0<a_k<1$.)

edited Apr 13 '17 at 12:21

Community

1

answered Dec 01 '15 at 13:31

Martin Sleziak

53,687

Yes, this is indeed a good and fast way to solve it, but I'm looking for a way to solve it without using sum series since we "didn't learn them yet". – user1640736 Dec 01 '15 at 13:35
@user1640736 Knowing that $\lim\limits_{n\to\infty} (1+\frac12+\dots+\frac1n)=+\infty$ would suffice for this, but this is just a different way of writing $\sum \frac1k = +\infty$. – Martin Sleziak Dec 01 '15 at 13:41
the other way to see this is to take logs which turns it into a normal sum, then use first order taylor expansion, to show the log goes to $-\infty$, so the original series goes to 0. – Alan Dec 01 '15 at 14:26

score 8 · Answer 2 · answered Dec 01 '15 at 14:16

After some research and thinking I have realised the following :

We can observe that : $$(2K-1)*(2K+1) < (2K)^2$$

Now, if we give values to K ( from 1 to n ) and the multiply all of them together we get : $$ 1 * 3^2 * 5^2 * ... * (2n-1)^2 * (2n+1) < 2^2 * 4^2 * ... * (2n)^2 $$

Now by dividing with $2n+1$ and the right side plus taking the square root of the entire identiy we get :

$$ {\frac {(2n-1)!!}{(2n)!!}} < {\frac {1}{\sqrt{2n+1}}}$$

Since the second term has the limit 0 as n goes to infinity and the left side is positive we get that :

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

score 6 · Accepted Answer · answered May 09 '14 at 17:08

Since you are asking for an idea, then you can use property of double factorial. Click the given link, take a look equation $(9)$ and $(11)$. You will obtain $$ \frac{(2n-1)!!}{(2n)!!}=\frac{(2n)!}{(2^n\ n!)^2}=\frac{(2n)!}{4^n(n!)^2}, $$ then use Stirling's approximation $$ n!\sim\sqrt{2\pi n}\left(\frac ne\right)^n $$ as $n\to\infty$.

Enrico M. · Answer 4 · 2015-12-01T14:12:07.260

3

May this help a bit?

For positive integers, those relations hold:

$$(2n)!! = 2^n n!$$

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

But I guess that your case need the general $n$ not only integers..

EDIT DUE TO THE COMMENTS BELOW

Using the relationships above and the Stirling approximations below, we find:

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

Applying the Stirling approximations:

$$\frac{(2n)!}{2^{2n} (n!)^2} = \frac{\sqrt{2\pi 2n} \left(\frac{2n}{e}\right)^{2n}}{2^{2n}(2\pi n)\left(\frac{n}{e}\right)^{2n}} = \frac{1}{\sqrt{\pi n}}$$

Chic goes to zero as $n$ goes to $+\infty$

edited Dec 01 '15 at 14:12

answered Dec 01 '15 at 13:55

Enrico M.

26,114

We didn't study factorials from anything else but integers, also positive integers. Not sure if those formulas help but they sure are interesting! – user1640736 Dec 01 '15 at 13:58
@user1640736 Good! Then if you use them, and you use them with Stirling, namely:
$$(2n)! \approx 2\sqrt{\pi n}\left(\frac{2n}{e}\right)^{2n}$$

and

$$(n!)(n!) \approx 2^{2n}(2\pi n)\left(\frac{n}{e}\right)^{2n}$$

You will get at the end:

$$\frac{1}{\sqrt{\pi n}}$$

which tends to zero as $n$ goes to infinity!
– Enrico M. Dec 01 '15 at 14:02

score 3 · Answer 5 · answered May 09 '14 at 16:51

Using \begin{align} (2n-1)!! &= \frac{2^{n} \Gamma(n+1/2)}{\sqrt{\pi}} \\ (2n)!! &= 2^{n} n! \end{align} then \begin{align} \frac{(2n-1)!!}{(2n)!!} = \frac{\Gamma(n+1/2)}{\sqrt{\pi} \Gamma(n+1)}. \end{align} Now using the duplication formula, $\Gamma(2x) = (2\pi)^{-1/2} 2^{2x-1/2} \Gamma(x) \Gamma(x+1/2)$, this fraction becomes \begin{align} \frac{(2n-1)!!}{(2n)!!} = \frac{1}{4^{n}} \binom{2n}{n}. \end{align} Since $4^{n}$ grows faster than the binomial component it is then seen that \begin{align} \lim_{n \rightarrow \infty} \left[ \frac{(2n-1)!!}{(2n)!!} \right] = \lim_{n \rightarrow \infty} \left[ \frac{1}{4^{n}} \binom{2n}{n} \right] \rightarrow 0. \end{align}

The thing you just state without proof "since $4^n$ grows faster than the binomial component" is basically the same type of question as the one being asked... This is to me a refrashing and not a proof. — Winther, May 09 '14 at 16:57

score 3 · Answer 6 · answered May 09 '14 at 18:14

Consider $(2n)!$
$(2n)!$=$2n(2n-1)(2n-2).........2(1)$
=$[(2n-1)(2n-3).....1][(2n)(2n-2)(2n-4).....2]$
=$(2n-1)!!$$2^n$$n!$
which gives $(2n-1)!!=(2n)!$$/$$2^n$$n!$
$(2n)!!$$=$$(2n)(2n-2)........2$
=$[2n][2n-2][2n-4]....2$
=$2^n$$n!$

so $(2n-1)!!$$/$$(2n)!!$$=$$(2n)!!$/$4^n$$[$(n)!$]^2$
$(2n-1)!!$$/$$(2n)!!$$=$$2n\choose n$$4^n$
using Stirling's Approximation
$2n\choose n$$=$$(2n/e)^{2n}\sqrt{2\pi*2n}$/$(n/e)^n\sqrt{2\pi*n}$
$=$$4^n$$[(n/e)^{2n}2\sqrt{\pi*n}]$/$[(n/e)^{2n}{2\pi*n}]$
$=$$4^n$/$\sqrt{\pi*n}$

which gives

$(2n-1)!!$$/$$(2n)!!$$ = $1/$\sqrt{\pi*n}$

finaly the limit $n$$\longrightarrow$$\infty$

$=$$0$

score 2 · Answer 7 · answered Dec 01 '15 at 13:54

2

$\lim\limits_{n\to \infty}\dfrac{(2n-1)!!}{(2n)!!}=0\quad$ and $\quad\lim\limits_{n\to \infty}\dfrac{(2n+1)!!}{(2n)!!}=\infty.~$ Where it really gets interesting, however,

is when we attempt to evaluate their product. Their polar-opposite tendencies will cancel each

other out, yielding $~\lim\limits_{n\to \infty}\dfrac{(2n-1)!!}{(2n)!!}\cdot\dfrac{(2n+1)!!}{(2n)!!}=\dfrac2\pi~,~$ which is the famous Wallis product.

See also Basel problem for more information.

answered Dec 01 '15 at 13:54

Lucian

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This is awesome for my own knowledge but doesn't really help me solve the limit :) – user1640736 Dec 01 '15 at 13:57
@user1640736: This is only meant to supplement the other answer. I can't think of a better way than the one already presented. – Lucian Dec 01 '15 at 13:59

How to prove that $\lim_{n \to\infty} \frac{(2n-1)!!}{(2n)!!}=0$

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