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

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!

• Use the fact that if $n>3$, then $2n-1>7$. Feb 27 at 10:31
• Note that both your observations are wrong, since $f(n+1) = (2n+1)\times f(n)$, and that is also odd. Feb 27 at 10:58

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.
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 $$\sigma_{2n-1}>7^{n-6}\sigma_5$$