An Inequality: $\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot\ldots\cdot\frac{99}{100} < \frac{1}{10}$

A product of fractions $\displaystyle \frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot\ldots\cdot\frac{2n-1}{2n}$ is on the left-hand side of several inequalities: one with a beautiful proof, one that strengthens the former but is virtually impossible to prove, and a third, even stronger, with an elementary proof.

Try your hand with the simplest variation:


$\displaystyle \frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot\ldots\cdot\frac{99}{100} \lt\frac{1}{10}.$


|Contact| |Front page| |Contents| |Up|

Copyright © 1996-2018 Alexander Bogomolny

Denote the left-hand side of the inequality A:

$\displaystyle A = \frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot\ldots \cdot\frac{99}{100}.$

And introduce its nemesis $B$:

$\displaystyle B = \frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}\cdot\ldots\cdot\frac{98}{99}.$

Factor by factor, the fractions in $B$ exceed those in $A:$

$\displaystyle \frac{2}{3} \gt \frac{1}{2},$ $\displaystyle \frac{4}{5} \gt \frac{3}{4},\ldots,\frac{98}{99} \gt \frac{97}{98},$ $\displaystyle 1 \gt \frac{99}{100}.$

From this it follows that $A \lt B.$ Note that, due to the choice of $B,$ in the product $AB$ most of the terms cancel out: $\displaystyle AB = \frac{1}{100}.$ From here,

$\displaystyle A^{2} \lt AB = \frac{1}{100},$

which, with one additional step, proves (1).

This proof suggests that (1) is in fact just a special case of a more general inequality


$\displaystyle \frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot\ldots \cdot\frac{2n-1}{2n} \lt \frac{1}{\sqrt{2n}},$

whose proof is a slight modification of the above with $A$ and $B$ defined as

$\displaystyle A(n) = \frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot\ldots \cdot\frac{2n-1}{2n},\\ \displaystyle B(n) = \frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}\cdot\ldots \cdot\frac{2n-2}{2n-1}.$

As we shall see shortly, (1) and (2) are quite weak: $A(n)$ has a much better bound, viz.


$\displaystyle A(n) \lt\frac{1}{\sqrt{3n+1}}.$

(3) supplies an edifying curiosity. By itself, it is easily proved by mathematical induction. However, its weakened version


$\displaystyle A(n) \lt\frac{1}{\sqrt{3n}},$

as far as I know, does not submit to an inductive proof. Try it, by all means. (3) and (3') are often quoted as a pair of problems of which the harder one has a simpler proof.

Meanwhile here's a proof for (3).

|Contact| |Front page| |Contents| |Up|

Copyright © 1996-2018 Alexander Bogomolny

To remind,

$\displaystyle A(n) = \frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot\ldots \cdot\frac{2n-1}{2n}$

and we wish to prove (3): $\displaystyle A(n) \lt\frac{1}{\sqrt{3n+1}}.$ For $n = 1,$ we have

$\displaystyle A(1) = \frac{1}{2} = \frac{1}{\sqrt{3\cdot 1+1}}.$

But already for $n = 2,$

$\displaystyle A(2) = \frac{1}{2}\cdot\frac{3}{4} = \frac{3}{8} \lt\frac{1}{\sqrt{7}} = \frac{1}{\sqrt{3\cdot 2+1}},$

because upon squaring $\displaystyle \frac{9}{64} \lt\frac{1}{7},$ for $7\cdot 9 = 63 \lt 64.$ Thus let's proceed with the inductive step and assume that (3) holds for $n = k:$


$\displaystyle A(k) \lt\frac{1}{\sqrt{3k+1}}.$

We are going to prove that, for $n = k+1,$ (3) also holds


$\displaystyle A(k+1) \lt\frac{1}{\sqrt{3(k+1)+1}} =\frac{1}{\sqrt{3k+4}}.$

Since $\displaystyle A(k+1) = A(k)\cdot\frac{2k+1}{2k+2},$ (4) implies


$\displaystyle A(k+1) \lt\frac{2k+1}{2k+2}\cdot\frac{1}{\sqrt{3k+1}}.$

Now square the right hand side in (6):

$\displaystyle \begin{align} \left(\frac{2k+1}{2k+2}\cdot\frac{1}{\sqrt{3k+1}}\right)^{2}&= \frac{(2k+1)^{2}}{(2k+2)^{2}(3k+1)}\\ &= \frac{(2k+1)^{2}}{12k^{3} + 28k^{2} + 20k + 4}\\ &= \frac{(2k+1)^{2}}{(12k^{3} + 28k^{2} + 19k + 4) + k}\\ &= \frac{(2k+1)^{2}}{(2k+1)^{2}(3k+4) + k}\\ &\lt\frac{(2k+1)^{2}}{(2k+1)^{2}(3k+4)}\\ &= \frac{1}{3k+4}, \end{align}$

which is exactly the right-hand side of (5) and proves (6).

Curiously, a much weaker $\displaystyle A(n) \lt\frac{1}{\sqrt{n}}$ is still resistant to the inductive argument, whereas a stronger version $\displaystyle A(n) \lt\frac{1}{\sqrt{n + 1}}$ goes through without a hitch.

(There is another example where mathematical induction applies easily to a stronger inequality and does not seem to work for a weaker one.)


  1. A. Engel, Problem-Solving Strategies, Springer Verlag, 1998, p. 180
  2. D. Fomin,S. Genkin,I. Itenberg, Mathematical Circles (Russian Experience), AMS, 1996, p. 90
  3. S. Savchev, T. Andreescu, Mathematical Miniatures, MAA, 2003, p. 51
  4. D. O. Shklyarsky, N. N. Chentsov, I. M. Yaglom, Selected Problems and Theorems of Elementary Mathematics, v 1, Moscow, 1959. (In Russian)

|Contact| |Front page| |Contents| |Up|

Copyright © 1996-2018 Alexander Bogomolny