Showing multiplication inequality using induction

Question

Use induction to show that: $$\frac{1}{2}\frac{3}{4}\dots \frac{2n-1}{2n} < \frac{1}{\sqrt{3n+1}} $$ for $n > 1$.

Answer 1

Let's write this formally. For $n > 1$, $$ \prod_{k=1}^n \frac{2k-1}{2k} < \frac{1}{\sqrt{3n+1}} $$ Initial case: Substitute $n = 2$ and use a calculator.

Inductive hypothesis: $$ \prod_{k=1}^{n-1} \frac{2k-1}{2k} < \frac{1}{\sqrt{3n-2}} $$

Inductive step: $$ \prod_{k=1}^n \frac{2k-1}{2k} = \frac{2n-1}{2n} \prod_{k=1}^{n-1} \frac{2k-1}{2k} < \frac{2n-1}{2n}\frac{1}{\sqrt{3n-2}} $$ Flipping and throwing $(2n)/(2n-1)$ into the square root, (We need to show that the following is $\geq\sqrt{3n+1}$) $$ \frac{2n}{2n-1} \sqrt{3n-2} = \sqrt{\frac{(3n-2)(4n^2)}{(2n-1)^2}} = \sqrt{\frac{12n^3 - 8n^2}{4n^2 - 4n + 1}} $$ Preforming polynomial long division, the last term is equal to $$ \sqrt{3n+1 + \frac{n-1}{4n^2 - 4n + 1}} \geq \sqrt{3n+1} $$ Since $\frac{n-1}{4n^2 - 4n + 1} > 0$. Therefore, $$ \frac{2n-1}{2n}\frac{1}{\sqrt{3n-2}} < \frac{1}{\sqrt{3n+1}} $$