Show that $\frac{1\cdot 3\cdot 5\cdot \ldots \cdot (2n-1)}{1\cdot 2\cdot 3\cdot \ldots \cdot n}\leq 2^{n}$ for all $n\in\mathbb{N}$.

Question

Show that $$\frac{1\cdot 3\cdot 5\cdot \ldots \cdot (2n-1)}{1\cdot 2\cdot 3\cdot \ldots \cdot n}\leq 2^{n}\qquad (n\in \mathbb{N}).$$

I want to show the last step, that is, the inductive step. Assume that this equation is true for some $n=k$. Note that
$$\frac{1\cdot 3\cdot 5\cdot \ldots \cdot (2n-1)}{1\cdot 2\cdot 3\cdot \ldots \cdot n}=\prod_{i=1}^{n}\left [ 2-\frac{1}{i} \right ].$$ For the case $n=k+1$ would be $$\prod_{i=1}^{k+1}\left [ 2-\frac{1}{i} \right ]=\prod_{i=1}^{k}\left [ 2-\frac{1}{i} \right ]\left [ 2-\frac{1}{k+1} \right ]\leq 2^{k}\left [ 2-\frac{1}{k+1} \right ]=2^{k+1}-\frac{2^{k}}{k+1}.$$ I have to show that the last term is $\leq 2^{k+1}$. But it's not possible to show that $2^{k}/(k+1)\geq 0$.

Answer 1

Another approach: $$\frac{1\cdot 3\cdot 5\cdot \ldots \cdot (2n-1)}{1\cdot 2\cdot 3\cdot \ldots \cdot n}< \frac{2\cdot 4\cdot 6\cdot \ldots \cdot 2n}{1\cdot 2\cdot 3\cdot \ldots \cdot n}=\frac{2^n n!}{n!}=2^n$$

Answer 2

For the step from $n=k$ to $n=k+1$, we multiply by $\frac{2k+1}{k+1}$, which is less than $2$.

Answer 3

Clearly, $2r-1<2r$ $$\implies\prod_{r=1}^n\dfrac{2r-1}{2r}<1$$

$$\implies\prod_{r=1}^n(2r-1)<\prod_{r=1}^n(2r)=2^n\prod_{r=1}^nr$$