3

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}$

Ivan Di Liberti
  • 4,881
user123451241
  • 633
  • 1
    That is probability of exactly 50 heads in 100 tosses of a fair coin. Can be estimated by Normal function approximation. But I suppose you want a neat proof. – Maesumi Jan 09 '13 at 21:07
  • 7
    No self work shown, no ideas, tries...not even "please" written. – DonAntonio Jan 09 '13 at 21:22

3 Answers3

6

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 $A \times \frac {99}{100} < \frac {1}{10}$.

Calvin Lin
  • 68,864
2

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.

coffeemath
  • 29,884
  • 2
  • 31
  • 52
1

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 $$.01< 0.012564513...$$

Tim Monahan
  • 577
  • This is way too much brute force, and doesn't provide an approach to dealing with the general version. You didn't need the first 2/3's if all you're going to do is multiply out everything and evaluate with a calculator. – Calvin Lin Jan 10 '13 at 16:48
  • I should consider myself lucky you used a “general version” of “way too” and didn’t say that “this is $way^{50} too^{49}$ much brute force” then I would have had to do a brute force calculation to figure out what you really meant. Kidding aside, the point of my answer was to show the OP how to convert the odd and even multiplications to factorials and give in an answer at the same time, thus the useless 2/3’s as you say. The calculations on the LHS were not done with brute force but the resulting RHS I guess is. I am not sure what the general version solution is to this specific problem. – Tim Monahan Jan 10 '13 at 19:23