<s>[INST] prove $ \frac{1}{13}<\frac{1}{2}\cdot \frac{3}{4}\cdot \frac{5}{6}\cdot \cdots \cdot \frac{99}{100}<\frac{1}{12} $ Without the aid of a computer，how to prove 
$$ \frac{1}{13}<\frac{1}{2}\cdot \frac{3}{4}\cdot \frac{5}{6}\cdot \cdots \cdot \frac{99}{100}<\frac{1}{12} $$
 [/INST] When you multiply up and down in the product by $2 \cdot 4 \cdot 6 \ldots 50$, the product becomes
$$\frac{1}{2^{100}} \binom{100}{50} = \frac{1}{2^{100}} \frac{100!}{(50!)^2}$$
Use Stirling:
$$n! \sim \sqrt{2 \pi n} n^n e^{-n} \left ( 1+ \frac{1}{12 n}\right )$$
for large $n$.  We will see what that means: plug in this approximation into the binomial expression above for $n=50$; the result is
$$\frac{1}{2^{100}} \binom{100}{50} \approx \frac{1}{\sqrt{50 \pi}} \left ( 1- \frac{1}{400}\right )$$
Note that the product being about $\frac{1}{\sqrt{50 \pi}}$ is accurate to $1$ part in $400$.  Now, it is clear that $50 \pi \approx 157$ to within that margin of error.  As $157$ falls between $12^2=144$ and $13^2=169$, the assertion is true.
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