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.

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