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What is the answer of this limit?

$$ \frac{1\times3\times5\times\cdots}{2\times4\times6\times8\times\cdots} $$

I remember that it was something involving $\pi$.

How can I compute it?

in addition; how can i compute it's Sum Serie Limit?

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I feel like this is just going to $0$. You might actually mean $\frac{\pi}{2}=\frac{2\cdot2\cdot4\cdot4\cdot6\cdots}{1\cdot3\cdot3\cdot5\cdot5 \cdots }$ – vrugtehagel Mar 11 at 18:50
@vrugtehagel I think I've proved it $0$, but idk if i'm right.. – Nikunj Mar 11 at 18:52
@Nikunj, be sure to post it as answer if you think you've solved it! There will be lots of people reading your proof and if something's wrong, they'll most likely comment. – vrugtehagel Mar 11 at 18:53
$\displaystyle \frac{1\times3\times5\times7}{2\times4\times6\times8} = \frac{1\times3\times\cdots \times (2n-1)}{2\times4\times\cdots\times(2n)} \vphantom{\frac {}{\displaystyle\int}} $ when $n=4$. One must consider $\displaystyle\lim_{n\to\infty} \frac{1\times3\times\cdots \times (2n-1)}{2\times4\times\cdots\times(2n)}$ and not $\displaystyle \frac{\lim_{n\to\infty} 1\times3\times\cdots\times(2n-1)}{\lim_{n\to\infty}2\times4\times\cdots\times(2n‌​)} \vphantom{\frac {}{\displaystyle\int}} $. $\qquad$ – Michael Hardy Mar 11 at 18:54
@vrugtehagel could you check it? – Nikunj Mar 11 at 18:56
up vote 10 down vote accepted

If you take the numerator product up to $2n-1$ and the denominator up to $2n,$ you have exactly $$ \frac{(2n)!}{4^n (n!)^2} $$

By Stirling's approximation this is asymptotic to $$ \frac{1}{\sqrt {\pi n}} $$ and goes to $0$ slowly. If we were to add one more term to the numerator, that is $2n+1,$ the expression would go to $\infty$ slowly, as a constant times $\sqrt n.$

The only simple way to get a limit is to take the numerator up to $2n-1$ as before, but also multiply the numerator by a single factor of $\sqrt n.$ Then the limit would be $1/ \sqrt \pi$

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@SubhadeepDey, IF we multiply by an extra $\sqrt n,$ we get a nonzero finite limit. The original thing goes to zero slowly. – Will Jagy Mar 11 at 19:12
Thanks... it was very helpful – Bidgoli Mar 12 at 20:07

As has been pointed out, the limit is zero.

However you mention that you remember it's something involving $\pi$. What you're probably thinking of is something like the result

$$\lim_{n \to \infty} \sqrt{n} {1 \cdot 3 \cdots (2n-1) \over 2 \cdot 4 \cdots 2n} = \frac 1{\sqrt{\pi}}.$$

Notice the additional factor of $\sqrt{n}$ in front. This means that your initial product, with $n$ factors on the top and $n$ factors on the bottom, is near $1/{\sqrt{\pi n}}$ when $n$ is larger. For example if $n = 100$ the product is about 0.05634 and $1/\sqrt{100\pi} \approx 0.05642$.

The limit quoted above can be proved using Stirling's formula,

$$\lim_{n \to \infty} {n! \over \sqrt{2\pi n} (n/e)^n} = 1, $$

if you rewrite the product as $(2n)!/(2^n n!)^2$.

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Thanks to you two! Dear Michael s! – Bidgoli Mar 12 at 20:08

Here you need to find $\lim A_n$, where $$A_n= \frac{1\times3\times5\times\cdots\times(2n-1)}{2\times4\times6\times8\times\cdots\times (2n)}.$$

Now note that, for $n\gt1$, $2n-1\gt n$. So, in the numerator, you get $$\frac{1\times 2\times 3\dots\times n}{2\times4\times 6\times\dots\times 2n}\lt\frac{1\times3\times5\times\cdots\times(2n-1)}{2\times4\times6\times8\times\cdots\times (2n)}\\\implies\frac1{2^n}\lt \frac{1\times3\times5\times\cdots\times(2n-1)}{2\times4\times6\times8\times\cdots\times (2n)}$$

Now, you use the identity $$\frac n{n+1}\lt\frac {n+1}{n+2}$$ and define a sequence $$B_n=\frac 23\times \frac 45\times\dots \times \frac {2n}{2n+1} .$$

So, now you have $$A_n\lt B_n\\\implies (A_n)^2\lt A_nB_n=\frac 1{2n+1}\\\implies A_n\lt \frac 1{\sqrt{2n+1}}.$$

So, you have $$\frac 1{2^n}\lt A_n\lt \frac 1{\sqrt{2n+1}}\\\implies\lim \frac 1{2^n}\lt\lim A_n\lt\lim \frac 1{\sqrt{2n+1}}.$$

So, by sandwitch theorem, $$\lim A_n=0.$$

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What's wrong with this? – Subhadeep Dey Mar 11 at 19:06
Dear downvoter, are you going to explain the reason for downvoting? – Subhadeep Dey Mar 11 at 19:18
I didn't downvote, but the first half of the proof is unnecessary. For the sandwich theorem, all you need is $0 < A_n < 1/\sqrt{2n+1}$, and $0 < A_n$ is obvious. – alephzero Mar 11 at 23:51
@alephzero, yes, I know. I have just showed that there are several ways to get the lower sequence. – Subhadeep Dey Mar 12 at 7:49
JUST BRILLIANT... – Bidgoli Mar 12 at 20:06

To paraphrase Wikipedia: if $a_n$ is a sequence of complex numbers such that $\sum_{n\geq 1} |a_n|^2 < \infty$, then non-zero convergence of $\prod_{n\geq 1}(1+a_n)$ is equivalent to convergence of $\sum_{n\geq 1} a_n$.

Letting $a_n = -1/2n$, we can see that, because the harmonic series diverges to infinity, your product converges to zero.

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If $0\leq a_n<1$ for all $n,$ then $$\prod_{n=1}^{\infty}(1-a_n)=0\iff \sum_{n=1}^{\infty}a_n=\infty.$$ Let $a_n=1/2 n.$ Then $\sum_{n=1}^{\infty}a_n=\infty.$ .

So $\quad \prod_{n=1}^{\infty}\frac {2 n-1}{2 n}=\prod_{n=1}^{\infty}(1-a_n)=0.$

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Proving the first assertion seems harder than the other proofs of the OP's result - unless it's a standard theorem that I've forgotten about. – alephzero Mar 12 at 0:00
@alephzero . It is a standard theorem,and not hard.It has a companion : If $a_n\geq0$ then $\prod_n(1+a_n)<\infty \iff \sum_n a_n <\infty.$ – user254665 Mar 12 at 0:15

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