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.
 A: for problem $3$ notice that the arithmetic average of the numbers $1,3,5,7,\dots,2n-1$ is $n$, and they are also $n$ terms.
So we have $n > \sqrt[n]{1\cdot 3 \cdot 5 \dots (2n-1)}$ by AM-GM. 
Raising to the $n$'th power yields the desired result.
A: 1) is equivalent to
$\prod\limits_{k=1}^n\frac{2k-1}{2k}>\frac{1}{2^n}
$
or
$2^n
>\prod\limits_{k=1}^n\frac{2k}{2k-1}
$.
But
$\frac{2k}{2k-1}
=1+\frac{1}{2k-1}
\le 2
$
for $k \ge 1$
and
$\frac{2k}{2k-1}
=1+\frac{1}{2k-1}
\lt 2
$
for
$k \ge 2$
which makes this evident.
A: Notice that in problem #$1$ if you raise each side to the $n$th power, then it is equivalent to showing that the product of the $n$ factors of the form
$$\left(1-\frac{1}{2k}\right)\tag{1}$$
is greater than $\left(\frac{1}{2}\right)^n$. But that is clearly true since each factor in equation $(1)$ is greater than or equal to $\frac{1}{2}$.
A: 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. 
A: 1) For $k\ge2$,
$$
\frac{2k-1}{2k}\gt\frac12
$$
Therefore, for $n\ge2$
$$
\begin{align}
\left[\prod_{k=1}^n\frac{2k-1}{2k}\right]^{1/n}
&=\left[\frac12\prod_{k=2}^n\frac{2k-1}{2k}\right]^{1/n}\\
&\gt\left[\frac12\prod_{k=2}^n\frac12\right]^{1/n}\\
&=\left[\prod_{k=1}^n\frac12\right]^{1/n}\\[6pt]
&=\frac12
\end{align}
$$

2) Squaring and cross-multiplication show that for $k\ge1$
$$
\frac{2k-1}{2k}\lt\sqrt{\frac{k-\frac12}{k+\frac12}}
$$
Therefore,
$$
\begin{align}
\prod_{k=1}^n\frac{2k-1}{2k}
&\lt\prod_{k=1}^n\sqrt{\frac{k-\frac12}{k+\frac12}}\\
&=\sqrt{\frac{\frac12}{n+\frac12}}\\
&=\frac1{\sqrt{2n+1}}
\end{align}
$$

3) The AM-GM says
$$
\begin{align}
\left(\prod_{k=1}^n(2k-1)\right)^{\large\frac1n}
&\le\frac1n\sum_{k=1}^n(2k-1)\\[6pt]
&=n
\end{align}
$$
