Mathematical Induction Problem

Prove that for every natural $n$, it is inequality:

$\frac { 1 }{ 2 } \times \frac { 3 }{ 4 } \times \frac { 5 }{ 6 } ...\frac { 2n-1 }{ 2n } \le \frac { 1 }{ \sqrt { 3n+1 } }$

#MathematicalInduction #Inequality

Easy Math Editor

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
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Comments

I am not sure what exactly I am supposed to do here, even though I have been on Brilliant for a year I haven't been regular. So, I hope I am doing the right thing.

$P(1)$ is true. Assume $P(n)$ to be true.

$P\left(n + 1\right) \implies \frac 1 2 \dots \frac {2n-1}{2n} \cdot \frac{2n + 1}{2n + 2} \leq \frac{1}{\sqrt{3n + 4}}$ From $P(n)$ , $\frac{1}{\sqrt{3n + 1}} \cdot \frac{2n + 1}{2n + 2} \leq \frac{1}{\sqrt{3n + 4}}$ After squaring and cross-multiplying, $\implies \left( 3n + 4 \right)\left(4n^2 + 4n + 1\right) \leq 4\left(n^2 + 2n +1\right)\left(3n + 1\right)$

I hope you can draw the conclusion from here.

Note: This is actually a very standard problem and can be found in almost every math contest book (In case, anyone is interested.)

Ishaan Singh - 7 years, 2 months ago

What is $4(n^2 + 2n + 1)(3n + 1) - (4n^2 + 4n + 1)(3n + 4)$ ? You should get something very simple.

Jon Haussmann - 7 years, 2 months ago

First of all, we can prove that " $\frac{2n+1}{2n+2} \leq \frac{\sqrt{3n+1}}{\sqrt{3n+4}}$ " is true.

Prove) Expansion the above inequality, we get $(4n^2+4n+1)(3n+4) \leq (4n^2+8n+4)(3n+1)$ . clean up this inequality, it become $0 \leq n$ . this is alway true when n is natural number.

if P(n) to be true and " $\frac{2n+1}{2n+2} \leq \frac{\sqrt{3n+1}}{\sqrt{3n+4}}$ " is also true, then $\frac{1}{2} \times \frac{3}{4} \times \frac{5}{6} \times \dots \times \frac{2n-1}{2n} \times \frac{2n+1}{2n+2} \leq \frac{1}{\sqrt{3n+1}} \times \frac{\sqrt{3n+1}}{\sqrt{3n+4}}$ ". so P(n+1) is also true.

P(1) is true, If P(n) is true, P(n+1) also true. So for every n, this inequality is true.

Sung Moo Hong - 7 years, 2 months ago

Nice question.

Edit: There is a easy proof for a looser bound go $\frac{1} { \sqrt{2n+1} }$ that doesn't use induction.

Edit: The RHS can be tightened to $\frac{ 1 } { \sqrt { \alpha n+ \beta} }$ for some constant $\alpha < \pi$ and $\beta$ . However, this requires more work.

Calvin Lin Staff - 7 years, 2 months ago

How is that tightening? $\frac{1}{\sqrt{2n + 1}}$ is greater than $\frac{1}{\sqrt{3n + 1}}$ .

Incidentally, $\frac{1}{2} \cdot \frac{3}{4} \dotsm \frac{2n - 1}{2n} = \frac{(2n)!}{4^n (n!)^2}.$ By Stirling's approximation, $\frac{(2n)!}{4^n (n!)^2} \approx \frac{1}{\sqrt{\pi n}}.$

Jon Haussmann - 7 years, 2 months ago

Oopps, wasn't thinking. Let me down vote myself.

Indeed, I wanted a coefficient of $\pi$ instead.

Calvin Lin Staff - 7 years, 2 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$