Does the sequence $a_n = \frac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{2 \cdot 4 \cdot 6 \cdots (2n)}$ converge?

Question

Does the sequence $a_n = \frac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{2 \cdot 4 \cdot 6 \cdots (2n)}$ converge?

This is a homework question for an analysis class.

Firstly, I know the sequence is comprised of odd numbers up to $2n-1$ over even numbers up to $2n$.

Secondly, I know the sequence decreases. I'm guessing it'll converge, but not specifically to $0$.

Where I am stuck is simplifying the series to something easier to work with. Any help would be greatly appreciated!

Edit: solved, thank you everyone. I ended up doing what was first suggested, not finding the actual limit but just proving the sequence converges. Thanks again!

Answer 1

If you're simply trying to show it converges and not concerned with what it actually converges to, the problem is quite trivial.

Since $2n-1 < 2n$ and $\frac{1}{2}$ is the first term, then $0 < x_{n} < 1$ for all $n$ so the sequence $\{x_{n}\}$ is a bounded sequence. Also notice that $x_{n+1} = x_{n}\frac{2n+1}{2n+2} < x_{n}$, then $\{x_{n}\}$ is monotone decreasing. So by the Monotone Convergence Theorem, the sequence converges.

Answer 2

If you know the Stirling formula, you can rewrite your sequence as

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

So, with Stirling formula, it's equivalent to

$$a_n \sim \frac{\sqrt{4n\pi}\left( \frac{2n}{e} \right)^{2n}}{2^{2n}\left(\sqrt{2n\pi}\left( \frac{n}{e} \right)^{n} \right)^2} = \frac{1}{\sqrt{n\pi}}$$

So it converge to $0$

Answer 3

$$a^{ 2 }_{ n }={ \left( \frac { 1\cdot 3\cdot 5\cdot ...\cdot (2n-1) }{ 2\cdot 4\cdot 6\cdot ...\cdot (2n) } \right) }^{ 2 }=\frac { { 1 }^{ 2 }\cdot 3^{ 2 }\cdot 5^{ 2 }\cdot ...\cdot (2n-1)^{ 2 } }{ 2^{ 2 }\cdot 4^{ 2 }\cdot 6^{ 2 }\cdot ...\cdot (2n)^{ 2 } } =\\ =\frac { 1\cdot 3 }{ { 2 }^{ 2 } } \cdot \frac { 3\cdot 5 }{ 4^{ 2 } } \cdot \frac { \left( 2n-1 \right) \left( 2n+1 \right) }{ \left( 2n \right) ^{ 2 } } \cdot \frac { 1 }{ 2n+1 } <\frac { 1 }{ 2n+1 } \\ \\ $$ for every $n$ $ \epsilon$ $ \mathbb{N} $ $0<{ a }_{ n }<\frac { 1 }{ \sqrt { 2n+1 } } \quad $ so it converges and

$$ \lim _{ n\rightarrow \infty }{ { a }_{ n } } =0$$

Answer 4

Rewrite the sequence as $$u_n=\frac{(2n)!}{(2\cdot 4\dotsm 2n)^2}=\frac{(2n)!}{2^{2n}(n!)^2}$$ and use Stirling's formula: $$u_n\sim\frac{\sqrt{4\pi n}\Bigl(\dfrac{2n}{\mathrm e}\Bigr)^{\!2n}}{2^{2n}2\pi n\Bigl(\dfrac{n}{\mathrm e}\Bigr)^{\!2n}}=\frac1{\sqrt{\pi n}}\to 0.$$