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Without the aid of a computer,how to prove

$$ \frac{1}{13}<\frac{1}{2}\cdot \frac{3}{4}\cdot \frac{5}{6}\cdot \cdots \cdot \frac{99}{100}<\frac{1}{12} $$

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Note that the middle quantity is really $\frac{100!}{2^{100} (50!)^2} = \frac{1}{2^{100}} {100 \choose 50}$. Then, this amounts to estimating the central binomial coefficient which is done, for example, here. This doesn't quite answer your question, since the bounds given are $ \frac{1}{2 \sqrt{50}}$ and $\frac{1}{2 \sqrt{51}}$, which are slightly worse than $\frac{1}{13}$ and $\frac{1}{12}$. – JavaMan Mar 6 '13 at 6:14
up vote 5 down vote accepted

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.

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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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"It is clear that..." No, it is not, unless you explain how the error depends on $n$. From what you wrote, it could well be that the two expressions are too far apart for the approximation to be useful, simply because $n=100$ is not "large" enough. – Andrés E. Caicedo Mar 6 '13 at 6:41
@AndresCaicedo: error on top is $(1+1/(24 n))$; error on bottom is $(1+1/(12 n))^2 \approx 1+1/(6 n)$. Ratio of error is then $1-1/(8 n)$, which is where I got my number. But never mind: the "it is clear" remark referred to the fact that $|50 \pi-157|/(50 \pi) < 1/400$, not the approximation that spawned the $1/400$ error estimate. – Ron Gordon Mar 6 '13 at 6:52
No, that is not the error I am talking about. Instead, I meant $|n!-\mbox{ugly expression}|$. (To be clear: Bounds for this error term are known, and are small enough to apply in the current case. But this needs to be addressed.) – Andrés E. Caicedo Mar 6 '13 at 6:56
You mean $|n!-\text{ugly expression}|/n! < 1/(288 n^2)$? For $n=50$, this relative error is $1/720000$, which is utterly negligible here. – Ron Gordon Mar 6 '13 at 7:05

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