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)

Prove that $\prod\limits_{i=1}^n \frac{2i-1}{2i} \leq \frac{1}{\sqrt{3n+1}}$ for all $n \in \Bbb Z_+$

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:



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?

  • $\begingroup$ For me, your prove seems correct. $\endgroup$ Feb 7, 2020 at 3:30
  • 1
    $\begingroup$ 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. $\endgroup$ Feb 7, 2020 at 3:43

4 Answers 4


Your calculations are correct.

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

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


  • $\begingroup$ Cool trick man! +1 $\endgroup$
    – Paramanand Singh
    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

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


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