Solve this inequality $\prod_{i=1}^{50} \frac {2i-1}{2i} < \frac {1}{10}$ Prove that
$ \frac{1}{2}\cdot \frac{3}{4}\cdot \frac{5}{6}\cdot \frac{7}{8}\cdot \frac{9}{10}\cdot \frac{11}{12}\cdot \frac{13}{14}...\cdot \frac{91}{92}\cdot \frac{93}{94}\cdot \frac{95}{96}\frac{97}{98}\cdot \frac{99}{100} <\frac{1}{10}$
 A: This is a standard / common problem.
Hint: 
$$\prod_{i=1}^{98} \frac {i}{i+1} = \frac {1}{99}$$
$$\sqrt{\frac {1}{99}} \times \frac {99}{100} = \frac {\sqrt{99}}{100} < \frac {1}{10}$$

Hint elaborated slightly:
Let $A = \prod_{i=1}^{49} \frac {2i-1}{2i}$ and $B= \prod_{i=1}^{49} \frac {2i}{2i+1}$. It is clear that $A<B$. 
The question wants us to show that $A \times \frac {99}{100} < \frac {1}{10}$.
A: The product of the first 31 fractions is $0.100923...$ while the product of the first 32 fractions is $0.099346...$, so it looks like you only need the first 32 of the 50 fractions to get below $1/10$. Using more fractions (all less than 1) will only make that smaller.
A: Multiply the numerators of the RHS and LHS by $49!2^{49}$. This converts the numerator on the LHS to $99!$. We then have $$\frac{99!}{2\cdot4\cdot8...\cdot100}<\frac{49!2^{49}}{10}$$ Next multiply the denominators on the RHS and LHS by $$\frac{100!}{50!2^{50}}$$ The formula to convert the above multiplications of even integers to factorials is $$\frac{(2n)!}{n!2^n}$$ where $n$ is the number of terms in the multiplication sequence. See OEIS A001147 for formula and sequence of multipliers to convert the sequential multiplications of even integers to factorials. This converts the denominator on the LHS to $100!$ to give $$\frac{99!}{100!}<\frac{2^{99}50!49!}{100!10}$$ The final result is $$\frac{1}{100}<\frac{19807040628566084398385987584}{1576427258524440520856445269625}$$ or $$.01< 0.012564513...$$
