# What is the non-inductive proof of this inequality? [duplicate]

$$\dfrac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{2 \cdot 4 \cdot 6 \cdots (2n)} < \dfrac{1}{\sqrt{3n+1}}.$$

However I've non-inductive proof of $\dfrac{1 \cdot 3 \cdot 5 \cdots(2n-1)}{2 \cdot 4 \cdot 6 \cdots(2n)} < \dfrac{1}{\sqrt{2n+1}}$, but I can't prove it for $3n+1$. It is obvious is to see $\dfrac{1 \cdot 3 \cdot 5 \cdots(2n-1)}{2 \cdot 4 \cdot 6 \cdots(2n)} < \dfrac{1}{\sqrt{3n+1}}$ is stronger inequality.

Let $S=\dfrac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{2 \cdot 4 \cdot 6 \cdots (2n)} < \dfrac{2 \cdot 4 \cdot 6 \cdots(2n)}{3 \cdot 5 \cdot 7 \cdots(2n+1)}$ [$\frac{1}{2}<\frac{2}{3}$ and so on.]

So, $S<\dfrac{1}{S(2n+1)}$ Implies, $S<\dfrac{1}{\sqrt{2n+1}}$

## marked as duplicate by Macavity, A.D, user147263, David K, DidJul 13 '15 at 22:17

• Have you tried using Stirling's approximation ? – Lucian Jul 10 '15 at 23:46
• no, I don't know such higher mathematics, this is basic non inequality question – Hardey Pandya Jul 10 '15 at 23:50
• This does not seem true for $n=1$ – TomGrubb Jul 11 '15 at 0:46
• @BolzWeir I was mistaken, sorry. You are right. (To be a stickler, the inequalities in the question should still be changed to weak inequalities though if we want it to hold for $n=1$) – TomGrubb Jul 11 '15 at 1:46
• – Macavity Jul 11 '15 at 3:36

$$S = \prod_{k=1}^n \frac{2k - 1}{2k} = \prod_{k=1}^n \frac{(2k)(2k - 1)}{(2k)^2} = \frac{(2n)!}{2^{2n}(n!)^2} = \frac{1}{4^n} {2n \choose n}$$
According to Wikipedia, ${2n \choose n} \le \frac{4^n}{\sqrt{3n + 1}}$