Prove the inequality $\frac{1}{1999}<\frac{1\cdot3\cdot5\cdot7\cdots1997}{2\cdot4\cdot6\cdot8\cdots1998}<\frac{1}{44}$ prove this inequality.
$\dfrac{1}{1999}<\dfrac{1\cdot3\cdot5\cdot7\cdots1997}{2\cdot4\cdot6\cdot8\cdots1998}<\dfrac{1}{44}$
I have tried to convert this series in factorial form. I am not getting what to do with this type of numbers $44$ and $1999$.
 A: Hint: Left inequality: Increase each term in the denominator by 1. Mass cancellation occurs.
Hint: Right inequality: Use the fact that $ (n-1)(n+1) < n^2$, show that
$$ A = \frac{ 1 \times 3 \times 5 \times \ldots 1997 } { 2 \times 4 \times 6 \times \ldots 1998 } < \frac{ 2 \times 4 \times 6 \times \ldots \times 1998} { 3 \times 5 \times \ldots \times 1999} = B.$$
Then, $ A^2 < AB = \frac{1}{1999}$
A: Hint:  $1\cdot3\cdot5\cdot7\cdots1997=\frac {1997!}{998!2^{998}}$, so
 $\dfrac{1\cdot3\cdot5\cdot7\dots1997}{2\cdot4\cdot6\cdot8\dots 1998}=\dfrac{(1\cdot3\cdot5\cdot7\dots1997)^2}{1998!}=\dfrac {(1997!)^2}{998!^22^{998}1998!}$  
Now Stirling's approximation should help  
Added: you can do the right by induction.  We want to prove that the expression with highest factor in the denominator $n$ is less than $\frac 1{\sqrt n}$.  Note that $\frac 12 \lt \frac 1{\sqrt 2}$.  Assume $\dfrac{1\cdot3\cdot5\cdot7\dots(k-1)}{2\cdot4\cdot6\cdot8\dots k}\lt \frac 1{\sqrt k}$  Then to get to $k+2$ we multiply on the left by $\frac {k+1}{k+2}$ and on the right by $\frac {\sqrt k}{\sqrt {k+2}}$
A: For the left inequality, take the 1998 on the bottom of the fraction and move it to the left, you now have:
$\frac 1 {1998} * \frac 32 * \frac 54 * \frac 76$...
Since $\frac 1 {1999} < \frac 1 {1998}$ and the right side is now being multiplied by a bunch of fractions greater than 1, the inequality still holds.
Working on the right inequality, but I think this is the easiest way for the left inequality.
A: Just to expand on the comment I made, let $X = \dfrac{1\cdot3\cdot5\cdot7\cdots1997}{2\cdot4\cdot6\cdot8\cdots1998}$.  Then the right inequality is equivalent to showing
$$ \frac1{X^2} > 44^2 \iff \frac{2^2}{1 \cdot 3} \cdot \frac{4^2}{3 \cdot 5}\cdot \frac{6^2}{5 \cdot 7} \cdots \frac{1998^2}{1997 \cdot 1999}> \frac{44^2}{1999}$$
$$\iff  \prod_{k=1}^{999} {\frac{(2k)^2}{(2k)^2-1}} > \frac{1936}{1999}$$
As all the factors on the left are greater than $1$ the inequality is evident.
For the left inequality, as the first comment mentions, simply multiply by $1999$ to get
$$\iff 1 < 1 \cdot \frac32 \cdot \frac54 \cdots\frac{1999}{1998}$$
which is obvious.
