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# Mathematical Induction Problem

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

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

Note by Carlos E. C. do Nascimento
3 years ago

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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.) · 3 years ago

What is $$4(n^2 + 2n + 1)(3n + 1) - (4n^2 + 4n + 1)(3n + 4)$$? You should get something very simple. · 3 years 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. · 3 years 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. Staff · 3 years 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, $\frac{(2n)!}{4^n (n!)^2} \approx \frac{1}{\sqrt{\pi n}}.$ · 3 years ago

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

Indeed, I wanted a coefficient of $$\pi$$ instead. Staff · 3 years ago