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 
$$ \frac{1}{13}<\frac{1}{2}\cdot \frac{3}{4}\cdot \frac{5}{6}\cdot \cdots \cdot \frac{99}{100}<\frac{1}{12} $$
 A: Another approach, perhaps simpler:  
Let $X = \frac{1}{2}\cdot \frac{3}{4}\cdot \frac{5}{6}\cdot \cdots \cdot \frac{99}{100}$  
We need to show $\dfrac{1}{13} < X < \dfrac{1}{12}$, which is equivalent to showing $144 < \dfrac{1}{X^2} < 169$, and we shall proceed to do this.  
For the lower bound, note that we need to show: $$ \frac{2^2}{1} \cdot \frac{4^2}{3^2}\cdot \frac{6^2}{5^2} \cdots \frac{100^2}{99^2} > 144$$
$$\Leftrightarrow  \frac{2^2}{1 \cdot 3} \cdot \frac{4^2}{3 \cdot 5}\cdot \frac{6^2}{5 \cdot 7} \cdots \frac{100^2}{99 \cdot 101} > \frac{144}{101} $$
$$\Leftrightarrow  \prod_{k=1}^{50} {\frac{(2k)^2}{(2k)^2-1}} > \frac{144}{101} $$
Now all the factors in the LHS are higher than $1$, and converge fairly fast to $1$, so taking the first three terms, we have $\dfrac{4}{3} \cdot \dfrac{16}{15} \cdot \dfrac{36}{35} = \dfrac {256}{175}$ which is easily verified to be higher than $\dfrac{144}{101}$.  Hence $X < \dfrac{1}{12}$ is established.  
For the other part viz. $\dfrac{1}{X^2} < 169$ the approach is exactly similar, you should get an equivalent inequality of form:
$$ \prod_{k=1}^{49} \frac{2k (2k+2)}{(2k+1)^2} < \frac{169}{200}$$  
Similarly, each factor now is less than $1$ and convergence is rapid.  So taking the first three terms, we have $\dfrac{1}{X^2} < \dfrac{1024}{1225}$ which is verifiably less than $\dfrac{169}{200} = 0.845$. Hence $\dfrac{1}{13} < X$ is also established.
A: 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.
