Proof without induction of the inequalities related to product $\prod\limits_{k=1}^n \frac{2k-1}{2k}$ How do you prove the following without induction:
1)$\prod\limits_{k=1}^n\left(\frac{2k-1}{2k}\right)^{\frac{1}{n}}>\frac{1}{2}$
2)$\prod\limits_{k=1}^n \frac{2k-1}{2k}<\frac{1}{\sqrt{2n+1}}$
3)$\prod\limits_{k=1}^n2k-1<n^n$
I think AM-GM-HM inequality is the way, but am unable to proceed. Any ideas. Thanks beforehand.

You may notice that the product appearing in 1) and 2) is related to binomial coeffients. 
$$
\frac{1\cdot3\cdots(2n-1)}{2\cdot4\cdots(2n)}=
\frac{(2n)!}{(2\cdot4\cdots(2n))^2}=
\frac{(2n)!}{(2^n n!)^2}=
\frac1{4^n} \binom{2n}n
$$
(See also here.)
So the inequalities in 1) and 2) are in fact equivalent to 
$$2^n < \binom{2n}n < \frac{4^n}{\sqrt{2n+1}}.$$
There already are several questions on this site about similar (or even stronger) inequalities for the central binomial coefficient. 
A sample of such questions (ordered by id)


*

*Elementary central binomial coefficient estimates

*Prove that $2^n < \binom{2n}{n} < 2^{2n}$

*To prove the inequality:- $\frac{4^m}{2\sqrt{m}}\le\binom{2m}{m}\le\frac{4^m}{\sqrt{2m+1}}$
You can probably find more using internal search, Google or Approach0. Or simply by checking the questions tagged binomial-coefficients+inequality. 
