How to prove $n^n \gt 1\cdot3\cdot.....(2n-1) $

Question

I am stuck with the following problem :

prove that : For $$n \ge 2, n^n \gt 1\cdot3\cdot.....\cdot (2n-1) $$

Can someone explain it as even though I tried it it was not fruitful and all my efforts went in vain.. Thanks in advance for your time.

Thanks a lot for the numerous responds..After posting the problem, I just realised that it can be done by using A.M. > G.M. as done by Michael Rozenberg.

Answer 1

I think you mean the following.

For all natural $n\geq2$ prove that: $$n^n>1\cdot3\cdot5\cdot...\cdot(2n-1).$$

If so, we can prove it by AM-GM:

$$n^2=1+3+5+...+2n-1>n\sqrt[n]{1\cdot3\cdot5\cdot...\cdot(2n-1)}$$ and we are done!

Answer 2

The original inequality is false: take $n=2$ then $2^2 = 4 < 6 = 3!$.

However, the reversed inequality $n^n < (2n-1)!$ holds for $n\ge 2$. It can be proven by induction: if it holds for $n$, then $$(n+1)^{n+1} = (n+1)^n (n+1) = n^n\left(1+\frac{1}{n}\right)^n \cdot (n+1) < (2n-1)! \cdot \left(1+\frac{1}{n}\right)^n \cdot (n+1)$$

so the only thing we prove is $$\left(1+\frac{1}{n}\right)^n \cdot (n+1)< 2n\cdot (2n+1).$$ The last inequality follows from $\left(1+\frac{1}{n}\right)^n \le e < 4$ and $4n+4\le 2n(2n+1)$ for $n\ge 2$.

Answer 3

Using the fact that $n!\sim \sqrt{2\pi n}(\frac{n}{e})^n$, we know that $$\infty=\lim_{n\to\infty}\frac{\sqrt{4\pi n}(\frac{2n}{e})^{2n}}{2n^{n+1}}=\lim_{n\to\infty}\frac{(2n)!}{2n*n^n}=\lim_{n\to\infty}\frac{(2n-1)!}{n^n}.$$

Thus, your claim is false and in fact $n^n<1*2*3*...*(2n-1)$.

Answer 4

There are already two proofs, but I think this one is more intuitive : $$ 1 \cdot 2 \cdot 3 \cdots (n-1)\cdot \underbrace{n \cdot (n + 1) \cdot (n+2) \cdots (2n-1)}_{n\ \text{terms that are all } \geq n} \geq [1\cdot 2 \cdot 3 \cdots (n-1)]\cdot n^n \overset{(*)}{>} n^n $$ $(*)$ Assuming $n > 2$

Answer 5

$n^n < 1\cdot 2\cdot3\cdot…(2n-1)$

Now $$1\cdot 2\dots \cdot(2n-1)=(1\cdot \overline{2n-1})\cdot(2\cdot \overline{2n-2})\cdots(k\cdot \overline{2n-k})\cdots n$$

Now for $1\leq k\leq 2n-1$ if, for some $k$, $k\cdot \overline{2n-k}\leq n$ then we will have $$2nk-k^2\leq n\Rightarrow k^2-2nk+n\geq 0\Rightarrow 4n^2-4n<0\Rightarrow n\leq 0$$

Thus $k\cdot \overline{2n-k}>n$

This follows the statement is true.

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I guess as @Michael Rozenberg said, this was not actually the question. But since his answer is elegant I am not giving the answer of the question what OP intended to say.(Although he/she did not edited until now, so...)