How to evaluate $\lim_{n\to\infty}\sqrt[n]{\frac{1\cdot3\cdot5\cdot\ldots\cdot(2n-1)}{2\cdot4\cdot6\cdot\ldots\cdot2n}}$

Im tempted to say that the limit of this sequence is 1 because infinite root of infinite number is close to 1 but maybe Im mising here something? What will be inside the root?
This is the sequence:

$$\lim_{n\to\infty}\sqrt[n]{\frac{1\cdot3\cdot5\cdot\ldots\cdot(2n-1)}{2\cdot4\cdot6\cdot\ldots\cdot2n}}$$

Solution 1:

Note that $$1\ge\frac{1\cdot3\cdot5\cdot\ldots\cdot(2n-1)}{2\cdot4\cdot6\cdot\ldots\cdot2n}=\frac{3\cdot5\cdot\ldots\cdot(2n-1)}{2\cdot4\cdot6\cdot\ldots\cdot(2n-2)}\cdot\frac1{2n}\ge\frac1{2n}$$ So the limit is $1$ by squeezing as $\sqrt[n]n\to1.$

Solution 2:

Note that $$ \left[\frac1{2n+1}\right]^{1/n}=\left[\frac13\frac35\frac57\cdots\frac{2n-1}{2n+1}\right]^{1/n}\le\left[\frac12\frac34\frac56\cdots\frac{2n-1}{2n}\right]^{1/n}\lt1 $$ Since $\lim\limits_{n\to\infty}(2n+1)^{1/n}=1$, the Squeeze Theorem says that $$ \lim_{n\to\infty}\left[\frac12\frac34\frac56\cdots\frac{2n-1}{2n}\right]^{1/n}=1 $$

Addendum

Since $1+x\le e^x$ for all $x\in\mathbb{R}$, we easily have $$ (1+\sqrt{n}/2)^2\le \left(e^{\sqrt{n}/2}\right)^2 $$ which implies that $$ 1+2n\le8e^{\sqrt{n}} $$ Therefore $$ 1\le(1+2n)^{1/n}\le8^{1/n}e^{1/\sqrt{n}} $$ By the Squeeze Theorem, $$ \lim_{n\to\infty}(2n+1)^{1/n}=1 $$

Solution 3:

If you insert the denominator into the numerator you have $$\frac{1\times 3\times 5\cdots \times(2n-1)}{2\times 4\times 6\cdots \times(2n)}=\frac{(2n)!}{4^n(n!)^2}$$ Now, use Stirling approximation $$m!=m^m \sqrt{2\pi m}e^{-m}$$