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Prove $\frac{1}{15}< \frac{1}{2}\times\frac{3}{4}\times\cdots\times\frac{99}{100}<\frac{1}{10}$

Things I have done: after trying many ways and failing, I reached the fact that$\left(\frac{1}{2}\times\frac{3}{4}\times\cdots\times\frac{99}{100}\right)^2<\left(\frac{1}{2}\times\frac{3}{4}\times\cdots\times\frac{99}{100}\right)\left(\frac{2}{3}\times\frac{4}{5}\times\cdots\times\frac{100}{101}\right)=\frac{1}{101}$

So $\frac{1}{2}\times\frac{3}{4}\times\cdots\times\frac{99}{100}<\frac{1}{10}$ is true. it remains to show $\frac{1}{15}< \frac{1}{2}\times\frac{3}{4}\times\cdots\times\frac{99}{100}$. I'm thinking of applying my approach on proving this part. something like this. $$\frac{1}{225}<\frac{1}{x}=\left(\frac{1}{2}\times\frac{3}{4}\times\cdots\times\frac{99}{100}\right) \times B<\left(\frac{1}{2}\times\frac{3}{4}\times\cdots\times\frac{99}{100}\right)^2$$

And another thing I'm curious about it,is there a way to approximate value of $\frac{1}{2}\times\frac{3}{4}\times\cdots\times\frac{99}{100}$ ?

user2838619
  • 3,120
  • Regarding your last question; the value you're looking for is the product $\displaystyle \prod_{n=1}^{50} \frac {2n-1}{2n}$. Mathematica gives an exact result of $\frac{12611418068195524166851562157}{158456325028528675187087900672}$ – UserX Sep 17 '14 at 11:10
  • @UserX,thanks.but i meant is there a way to approximate it without doing all multiplications? – user2838619 Sep 17 '14 at 11:15

2 Answers2

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Note that $$\frac{n-1}{n}\gt\frac{n-2}{n-1}$$ for $n\gt 1$. So, we have $$\begin{align}\left(\frac 12\times \frac 34\times \cdots\times \frac{99}{100}\right)^2&\gt \left(\frac 12\times \frac 34\times\frac 56\times \cdots\times \frac{99}{100}\right)\left(\color{red}{\frac 12}\times \frac 23\times \frac{4}{5}\times\cdots\times \frac{98}{99}\right)\\&=\frac{1}{200}\\&\gt\frac{1}{225}\\&=\frac{1}{15^2}\end{align}$$

mathlove
  • 139,939
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Don't you already have the answer?

$\left(\frac{1}{2}\times\frac{3}{4}\times\cdots\times\frac{99}{100}\right)^2>\frac{1}{2}\left(\frac{1}{2}\times\frac{3}{4}\times\cdots\times\frac{99}{100}\right)\left(\frac{2}{3}\times\frac{4}{5}\times\cdots\times\frac{98}{99}\right)=\frac{1}{200}$

Martigan
  • 5,844