Prove $\frac{1\cdot 3\cdots(2n-1)}{2\cdot 4 \cdots(2n)}<\frac{1}{\sqrt{2n+1}}$, $n\ge 1$ Prove $\frac{1\cdot 3\cdots(2n-1)}{2\cdot 4 \cdots(2n)}<\frac{1}{\sqrt{2n+1}}$, $n\ge 1$. 
I begin by letting $n=1$ then $\frac{1}{2}<\frac{1}{\sqrt{3}}$. Then assume  $\frac{1\cdot 3\cdots(2k-1)}{2\cdot 4 \cdots(2k)}<\frac{1}{\sqrt{2k+1}}$,  for some $k\ge 1$. Then if I multiply both sides by $\frac{(2(k+1)-1)}{2(k+1)}$ I get $\frac{1\cdot 3\cdots(2(k+1)-1)}{2\cdot 4 \cdots2(k+1)}<\frac{1}{\sqrt{2k+1}}\cdot \frac{(2(k+1)-1)}{2(k+1)}$. So I have what I need on the left hand side, but I'm not sure how to continue on the right hand side. I realized that if I show that $\frac{(2(k+1)-1)}{2(k+1)}<\frac{\sqrt{2k+1}}{\sqrt{2(k+1)+1}}$ then I would be done. This however leads to the same dilemma in that I can easily make the left the way I need, but not the right. So there must be a trick that I am missing.

First observe that
$$\frac{2n-1}{2n}<\frac{2n}{2n+1}\qquad\text{since }(2n-1)(2n+1)=4n^2-1<4n^2=(2n)(2n)$$
Then
\begin{align}
\left(\prod_{k=1}^{n}\frac{2k-1}{2k}\right)^2&<\left(\prod_{k=1}^{n}\frac{2k-1}{2k}\right)\left(\prod_{k=1}^{n}\frac{2k}{2k+1}\right)\\
&=\prod_{k=1}^{n}\frac{2k-1}{2k+1}\qquad\text{this product is "telescopic"}\\
&=\frac{1}{2n+1}
\end{align}
So
$$\prod_{k=1}^{n}\frac{2k-1}{2k}<\frac{1}{\sqrt{2n+1}}$$
