Question

Prove that $1/\left(2n\right)\le [\mathrm{1.3.5}\dots .(2n-1\left)\right]/(2.4\dots .2n)$whenever n is a positive integer.

Short answer

$1/\left(2n\right)\le [\mathrm{1.3.5}\dots .(2n-1\left)\right]/(2.4\dots .2n)$is true for all positive integers n.

Step-by-step solution

Step: 1

If n=1,

$\begin{array}{r}1/\left(2.1\right)\le [\mathrm{1.3.5}\dots \dots (2\left(1\right)-1\left)\right]/(2.4\dots 2(1\left)\right)\\ \frac{1}{2}=\frac{1}{2}\end{array}$

it is true for n=1.

Step: 2

Let P(k) be true.

$1/\left(2k\right)\le [\mathrm{1.3.5}\dots .(2k-1\left)\right]/(2.4\dots .2k)$

We need to prove that P(k+1) is true.

Step: 3

$\begin{array}{r}1/(2k+1)\le \frac{[1\cdot 3\cdot 5\dots (2k-1)\cdot 2(k+1)-1]}{(2\cdot 4\dots \cdot 2k)\cdot 2(k+1)}\\ \ge \frac{1}{2k}\cdot \frac{2(k+1)-1}{2(k+1)}\\ =\frac{2k+1}{4k(k+1)}\\ \ge \frac{1}{2(k+1)}\end{array}$

It is true for P(k+1) is true.