The Case of the Method of Mathematical Induction

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The Case of the Method of Mathematical Induction

One day, a teacher gave 10th grade students the following homework problem:

To prove, by attempting to use the method of mathematical induction, that for any natural number $n$ the inequality is true

$$P_n = \frac{1 \cdot 3 \cdot 5 \dots (2n-1)}{2 \cdot 4 \cdot 6 \dots 2n} < \frac{1}{\sqrt{n+1}}.$$

Kolya V. found the following solution at home;

a) For $n = 1$ the inequality is true, since

$$P_1 = \frac{1}{2} < \frac{1}{\sqrt{2}}.$$

6. Let us show that the validity of inequality $P_{n+1} < \frac{1}{\sqrt{n+2}}$ follows from condition $P_n < \frac{1}{\sqrt{n+1}}$. To do this it is enough to show that

$$\frac{P_{n+1}}{P_n} < \frac{\sqrt{n+1}}{\sqrt{n+2}},$$

or

$$\frac{2n+1}{2n+2} < \sqrt{\frac{n+1}{n+2}},$$

or

$$(2n+1)^2(n+2) < 4(n+1)^3.$$

or

$$4n^3 + 12n^2 + 9n + 2 < 4n^3 + 12n^2 + 12n + 4,$$

or

$$0 < 3n + 2,$$

which is obviously true. Thus, a fairly provable inequality follows from the principle of mathematical induction.

However, while telling this solution at school, Kolya forgot to put one in the denominator and began to prove a more “gross” inequality:

$$P_n < \frac{1}{\sqrt{n}}.$$

Here’s what he got:

a) For $n = 1$ the inequality is true, since

$$P_1 = \frac{1}{2} < \frac{1}{1}.$$

6. Let us show that condition $P_n < \frac{1}{\sqrt{n}}$ implies $P_{n+1} < \frac{1}{\sqrt{n+1}}$. To do this it is enough to show that

$$\frac{P_{n+1}}{P_n} < \frac{\sqrt{n}}{\sqrt{n+1}},$$

or

$$\frac{2n+1}{2n+2} < \sqrt{\frac{n}{n+1}},$$

or

$$(2n+1)^2(n+2) < 4(n+1)^3n,$$

or

$$4n^3 + 12n^2 + 9n + 2 < 4n^3 + 12n^2 + 12n + 4,$$

or

$$0 < 3n + 2,$$

which is obviously true. Thus, a fairly provable inequality follows from the principle of mathematical induction.

For those who understand this issue, we propose to prove two inequalities:

1. $P_n < \frac{1}{\sqrt{3,1n}}$.

It is interesting that in this inequality the coefficient 3.1 cannot be replaced by 3.2. The largest coefficient at which it remains true is π (the ratio of the circumference to the diameter).

2. $\frac{1}{2^3} + \frac{1}{3^3} + \dots + \frac{1}{n^3} < \frac{1}{4}$.