<s>[INST] 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 $1999$.
 [/INST] 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.
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