# Proof by induction:$\frac{3}{5}\cdot\frac{7}{9}\cdot\frac{11}{13}\cdots\frac{4n-1}{4n+1}<\sqrt{\frac{3}{4n+3}}$

In the very beginning I'm going to refer to similar posts with provided answers:

Induction Inequality Proof with Product Operator $\prod_{k=1}^{n} \frac{(2k-1)}{2k} \leq \frac{1}{\sqrt{3k+1}}$ (answered by Özgür Cem Birler)

I examined the solutions and tried to apply the methods used there to make sure whether I understand it or not. I'm concerned about the step of induction.

A task from an earlier exam at my university:

Prove by induction:

$$\frac{3}{5}\cdot\frac{7}{9}\cdot\frac{11}{13}\cdots\frac{4n-1}{4n+1}<\sqrt{\frac{3}{4n+3}}$$

Attempt:

rewritten: $$\prod_{i=1}^n\frac{4i-1}{4i+1}<\sqrt{\frac{3}{4n+3}}$$ $$(1)$$ base case: $$\tau(1)$$ $$\frac{3}{5}=\sqrt{\frac{4}{7}}<\sqrt{\frac{3}{7}}$$ $$(2)$$ assumption:

Let:$$\frac{3}{5}\cdot\frac{7}{9}\cdot\frac{11}{13}\cdots\frac{4n-1}{4n+1}<\sqrt{\frac{3}{4n+3}}$$ hold for some $$n\in\mathbb N$$

$$(3)$$ step: $$\tau(n+1)$$ $$\frac{4n+3}{4n+5}\cdot\prod_{i=1}^n\frac{4i-1}{4i+1}<\frac{4n+3}{4n+5}\cdot\sqrt{\frac{3}{4n+3}}=\frac{\sqrt{3(4n+3)}}{4n+5}<\sqrt{\frac{3}{4n+7}}$$ $$\frac{12n+9}{16n^2+40n+25}<\frac{3}{4n+7}\iff\frac{48n^2+120n+63-48n^2-120n-75}{\underbrace{(4n+5)^2(4n+7)}_{>0}}<0\iff-\frac{12}{(4n+5)^2(4n+7)}<0$$

Is this combined legitimate?

• For me, your prove seems correct. Feb 7, 2020 at 3:30
• Nice! This is the kind of proof where I think it could probably be written up in a much cleaner fashion. But everything is "technically correct" imo. Feb 7, 2020 at 3:43

But I thought it might be helpful also to mention another nice trick to handle such products:

• Set $$A = \frac 35 \cdot \frac 79 \cdots \frac{4n-1}{4n+1}$$
• Let $$B = \frac 57 \cdot \frac 9{11} \cdots \frac{4n+1}{4n+3}$$

It follows immediately

$$A < B \Rightarrow A^2 < AB = \frac 3{4n+3}$$

Done.

• Cool trick man! +1 Feb 9, 2020 at 9:01

I concur with everyone else that it's basically right. Everyone has their own style, but the following is how I would probably write up the "meat and potatoes" of the proof:

Let's start with some preliminary observations. Note that \begin{align} \frac{4n+3}{4n+5}\cdot\sqrt{\frac{3}{4n+3}} &=\frac{4n+3}{4n+5}\cdot\frac{\sqrt{3}} {\sqrt{4n+3}}\\[1em] &=\frac{\sqrt{4n+3}\cdot\sqrt{3}}{4n+5}\\[1em] &= \sqrt{\frac{(4n+3)(3)}{(4n+5)^2}}\\[1em] &= \sqrt{\frac{12n+9}{16n^2+40n+25}}. \end{align}

More concisely, we have $$\frac{4n+3}{4n+5}\cdot\sqrt{\frac{3}{4n+3}}=\sqrt{\frac{12n+9}{16n^2+40n+25}}.\tag{1}$$ As another observation, note that \begin{align} \frac{12x+9}{16x^2+40x+25} < \frac{3}{4x+7} &\iff \frac{12x+9}{16x^2+40x+25} - \frac{3}{4x+7} < 0\\[1em] &\iff \frac{(48x^2+120x+63)-(48x^2+120x+75)}{(4x+5)^2)(4x+7)}<0\\[1em] &\iff \frac{-12}{(4x+5)^2(4x+7)}<0\\[1em] &\iff \frac{12}{(4x+5)^2(4x+7)}>0\\[1em] &\iff x\in\Bigl(-\frac{7}{4},-\frac{5}{4}\Bigr)\cup\Bigl(-\frac{5}{4},\infty\Bigr). \end{align} More specifically, note that, for any natural number $$n$$, it follows from above that $$\frac{12n+9}{16n^2+40n+25} < \frac{3}{4n+7}.\tag{2}$$

Now your proof can flow very naturally: \begin{align} \prod_{i=1}^{k+1}\frac{4i-1}{4i+1} &=\prod_{i=1}^k\frac{4i-1}{4i+1}\cdot\frac{4(k+1)-1}{4(k+1)+1}\\[1em] &<\frac{4k+3}{4k+5}\cdot\sqrt{\frac{3}{4k+3}}\\[1em] &=\sqrt{\frac{12k+9}{16k^2+40k+25}} & \text{(by (1))}\\[1em] &<\sqrt{\frac{3}{4k+7}} & (\text{\sqrt{x} strictly increases and by (2)})\\[1em] &= \sqrt{\frac{3}{4(k+1)+3}}. & \text{(desired conclusion)} \end{align}

Maybe that's overdrawn and slightly verbose, but that's how I would write it up if I were doing it for a class.

Base case: $$n = 1$$

$$\frac {3}{5} < \sqrt {\frac {3}{7}}$$

Since both are positive, we can square both sides.

$$\frac {9}{25} < \frac {3}{7}$$

Cross multiply $$63 < 75$$

Inductive hypothesis

Suppose, $$\prod_\limits{i=1}^{n} \frac {4i-1}{4i+1} < \sqrt {\frac {3}{4n+3}}$$

We must show that when the hypothesis holds, $$\prod_\limits{i=1}^{n+1} \frac {4i-1}{4i+1} < \sqrt {\frac {3}{4n+7}}$$

$$\prod_\limits{i=1}^{n+1} \frac {4i-1}{4i+1} = \frac {4n+3}{4n+5}\prod_\limits{i=1}^{n} \frac {4i-1}{4i+1} < \frac {4n+3}{4n+5}\sqrt {\frac {3}{4n+3}}$$ based on the inductive hypothesis.

$$\frac {4n+3}{4n+5}\sqrt {\frac {3}{4n+3}} = \sqrt {\frac {3(4n+3)}{(4n+5)^2}}$$

$$(4n+5)^2 = (4n+3)(4n+7) + 4$$

$$\sqrt {\frac {3(4n+3)}{(4n+5)^2}} = \sqrt {\frac {3(4n+3)}{(4n+3)(4n+7) + 4}} < \sqrt {\frac {3(4n+3)}{(4n+3)(4n+7)}} = \sqrt {\frac {3}{4n+7}}$$

Which is what we were required to show.

It is correct.

Anyways you already ended the proof when you stated that

$$\frac{4n+3}{4n+5}\prod_{i=1}^n\frac{4i-1}{4i+1}$$ $$<$$ $$\sqrt{\frac{3}{4n+7}}$$