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Find:
$\displaystyle
\lim_{n \to \infty} \frac{1 \times 3 \times .. \times (2n-1)}{2\times 4 \times .. \times 2n}=
\lim_{n \to \infty} \frac{(2n)!}{(2\times 4 \times .. \times 2n)^2}=
\lim_{n \to \infty}\frac{(2n)!}{2^{2n} (n!)^2}
$
Now Stirlings formula for the factorial is:
$\displaystyle n! \sim \sqrt{2 \pi n}n^n e^{-n}$
so:
$\displaystyle
\frac{(2n)!}{2^{2n} (n!)^2}\sim
\frac{\sqrt{4 \pi n}(2n)^{2n}e^{-2n}}{2^{2n}(\sqrt{2 \pi n}n^n e^{-n})^2}=
\frac{\sqrt{2}}{\sqrt{2 \pi n}}
$
so:
$\displaystyle
\lim_{n \to \infty}\frac{(2n)!}{2^{2n} (n!)^2}=0
$
(I suppose to make this rigourus I should put in the error term for Stirlings formula
but I guess I will leave this tpo the reader)
RonL
Supporting evidence is in the attachment)
Let me do the second one, it looks more interesting.
I hope you are familar with the Euler-Wallis product formula.
Thus,
$\displaystyle \frac{\sin x}{x}=\left(1-\frac{x^2}{\pi^2}\right)\left( 1-\frac{x^2}{4\pi^2}\right)\left(1-\frac{x^2}{9\pi^2} \right)\cdot ...$
If,
$\displaystyle x=\pi/2$
We have,
$\displaystyle \frac{2}{\pi}=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{4^2}\right)\left(1-\frac{1}{6^2}\right)...$
Thus,
$\displaystyle \frac{2}{\pi}=\prod_{n=1}^{\infty}\frac{4n^2-1}{4n^2}$
Reciprocal time,
$\displaystyle \frac{\pi}{2}=\prod_{n=1}^{\infty} \frac{4n^2}{4n^2-1}$
Factor,
$\displaystyle \frac{\pi}{2}=\prod_{n=1}^{\infty} \frac{(2n)(2n)}{(2n-1)(2n+1)}=\frac{2\cdot 2\cdot 4\cdot 4\cdot...}{1\cdot 3\cdot 3\cdot 5\cdot 5...}$
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Now we do thy problem
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The limit of,
$\displaystyle \frac{1\cdot 3\cdot 5\cdot ... (2n-1)}{2\cdot 4\cdot 6\cdot ... (2n)}$
If you multiply the numerator and denominator by the missing terms in order to obtain the Euler-Wallis product you will have the expression above. But since $\displaystyle 2n+1$ is missing there will be an extra division so,
$\displaystyle \mbox{ Wallis Product }\cdot \frac{1}{2n+1}$
Now the Wallis product will converge to $\displaystyle \frac{2}{\pi}$ and $\displaystyle \frac{1}{2n+1}\to 0$ thus, the product of these two convergents tells us the limit is zero.
Note, I was extremely lazy in checking whether its square will converge to Euler-Wallis product because the last term is missing. So I might be wrong and the limit is zero.
The idea is that we are going to assume that the Dirichlet function is Riemann integrable then arrive at a contradiction.
A bit gappy, but then so is my "proof", but I suppose because its mine
I know how to fill those gaps (and that they do fill).
(I agree that this is more interesting, took some thought to deduce
what the limit must be - always an advantage before starting a proof)
RonL
Do not think that I came up with this full solution be realizing that the solutions for the sine functions form the infinite product. I have seen similar ones before, in fact, when I saw the evens and odd seperated by division a thought immediately occured to me, Wallis infinite product! I just manipulated it so that it worked. I think it would have been cooler is the user asked to find,
$\displaystyle \lim \frac{1\cdot 3\cdot ... \cdot (2n+1)}{2\cdot 4\cdot ... \cdot (2n)}$
In such a case the solution who have involved $\displaystyle \pi$! I was hoping for that badly, and assumed that was the case until I arrived that the conlusion that there is one more term in the demoninator which made a possibly amazing problem very boring.
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In fact one way Sirling's asympomatic approximation is derived is from what I said!