## 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:

(1)

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

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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

(2)

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

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.

(3)

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

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

(3')

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

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).

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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 $n = k:$

(4)

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

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

(5)

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

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

(6)

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

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.)

### References

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

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