Using $\sqrt{1-t}\leq 1-\frac t2$ to show that $\frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots2n}\geq\frac1{2\sqrt n}$

Question

I have a problem that tells me to use that $\sqrt{1-t}\leq 1-\frac t2$ for $t\in(0,1)$ to show by induction that $\frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots2n}\geq\frac1{2\sqrt n}$

So far I have shown that $\frac{1\cdot3\cdot5\cdots(2n-1)}{1\cdot2\cdot3\cdots n}\leq 2^n$ from a previous problem. I don't know if that can be used. But the first premise has to be used.

Daniel Fischer · Accepted Answer · 2014-06-12T10:03:56.693

8

Write the product

$$\prod_{k=1}^n \frac{2k-1}{2k} = \prod_{k=1}^n \left( 1 - \frac{1/k}{2}\right)$$

and use the given inequality for $k > 1$ to get the inequality

$$\prod_{k=1}^n \frac{2k-1}{2k} \geqslant \left(1-\frac{1}{2}\right)\prod_{k=2}^n \sqrt{1-\frac{1}{k}} = \frac{1}{2}\prod_{k=2}^n\sqrt{\frac{k-1}{k}}.$$

edited Jun 12 '14 at 10:03

answered Jun 11 '14 at 21:39

Daniel Fischer

206,697

I forgot to specify that the given inequality holds only for $t\in(0,1)$. – Alec Jun 11 '14 at 21:47
1

@Aleksander: since you are using the equality for $k > 1$, $1/k \in (0, 1)$. – Alex Wertheim Jun 11 '14 at 21:50
Sorry, but I'm not able to piece it together from this. I don't see how the given inequality fits into it. Especially with the $\sqrt{1-t}$ part. Any further pointers? – Alec Jun 12 '14 at 03:08
@Aleksander For $k > 1$, use $$1 - \frac{1/k}{2} \geqslant \sqrt{1-\frac{1}{k}}.$$ – Daniel Fischer Jun 12 '14 at 10:05

Using $\sqrt{1-t}\leq 1-\frac t2$ to show that $\frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots2n}\geq\frac1{2\sqrt n}$

1 Answers1

Linked

Related