Show that $\forall n\ge3,(2 n-1) \times(2 n-3) \times(2 n-5) \times \cdots \times 5 \times 3 \times 1\ge2 \times 7^{n-2}$

Question

Here's my question: Let $f(n)=(2 n-1) \times(2 n-3) \times(2 n-5) \times \cdots \times 5 \times 3 \times 1$, and I need to prove that for all $n\ge3,f(n)\ge2\times7^{n-2}$.

I have figured out some interesting points, but still I have no clue how to make use of them.

1. $f(n+1)=2n\times(2n-2)\times\dots\times4\times2=2^n\times n!$
2. $f(n+1)\times f(n)=(2n)(2n-1)\dots(3)(2)(1)=(2n)!$

Then everything stops here. I'm trying to make sense with the $7^{n-2}$ with the statement, but failed. So is there any other hints for this question? Thanks a lot for your help!

Answer 1

Just bound the terms this way: $$f(n)=\underbrace{(2n-1)}_{\geq 7}\underbrace{(2n-3)}_{\geq 7}\ldots\underbrace{(7)}_{\geq 7}(5)\underbrace{(3)}_{\geq 2}$$ and take into account that there are $n$ terms.

Answer 2

we put $\sigma_{2n-1}=(2n-1)!!$ for $n>3$ : We have $\sigma_5=9×7×5×3×2×1>2(7^{5-2})$
and $\sigma_7=9×\sigma_5>7\sigma_5$
By recurence we obtain $\sigma_{2n-1}>7^{n-6}\sigma_5$