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I am trying to prove an inequality by induction which is as follows:

$$\frac{(2n)!}{2^{2n}\cdot (n!)^2} \le \frac{1}{\sqrt{3n + 1}}$$

Base Case, i.e, for n =1, $$\frac{(2(1))!}{2^{2(1)}\cdot ((1)!)^2} \le \frac{1}{\sqrt{3(1) + 1}}$$ $$\implies \frac{2!}{2^{2}\cdot (1)^2} \le \frac{1}{\sqrt{3 + 1}}$$ $$\implies \frac{2}{4} \le \frac{1}{2}$$ $$\implies \frac{1}{2} \le \frac{1}{2}$$

which is true

Assume that the inequality is true for $n=m$ where $m \in \mathbb{N}$ , i.e, $$\frac{(2m)!}{2^{2m}\cdot (m!)^2} \le \frac{1}{\sqrt{3m + 1}}$$ $$\implies \quad \quad \quad\quad\quad\quad\quad \frac{1\cdot2\cdot3\cdot .....\cdot (2m-1)\cdot (2m)}{2^{2n}\cdot (1\cdot2\cdot3\cdot .....\cdot (m-1) \cdot m)^2} \le \frac{1}{\sqrt{3m + 1}} \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$$ $$\implies \quad \quad \quad \frac{1\cdot2\cdot3\cdot .....\cdot (2m-1)\cdot (2m) \cdot(2m+1)\cdot(2m +2)}{2^{2m}\cdot (1\cdot2\cdot3\cdot .....\cdot (m-1) \cdot m)^2\cdot(2m+2)\cdot (m+1)\cdot2} \le \frac{1}{\sqrt{3m+ 1}}\cdot \frac{(2m +1) }{(m+1)\cdot 2} \quad\quad\quad\quad\quad\quad\quad\quad$$ $$\implies \frac{(2m+2)!}{2^{2m +2}\cdot (1\cdot2\cdot3\cdot .....\cdot (m-1) \cdot m \cdot (m+1))^2 }\le \frac{1}{\sqrt{3m+ 1}}\cdot \frac{(2m +1) }{(m+1)\cdot 2}$$ $$\implies \frac{(2(m+1))!}{2^{2(m +1)}\cdot ((m+1)!)^2 }\le \frac{1}{\sqrt{3m+ 1}}\cdot \frac{(2m +1) }{(m+1)\cdot 2}$$ $$\implies \frac{(2(m+1))!}{2^{2(m +1)}\cdot ((m+1)!)^2 }\le \frac{1}{\sqrt{12m+ 4}}\cdot \frac{(2m +1) }{(m+1)}\quad \quad ......(i)$$$$

Now , I want $LHS$ to be smaller than or equalto $\frac{1}{\sqrt{3m+4}}$. For this I thought to prove that

$$\frac{1}{\sqrt{12m+ 4}}\cdot \frac{(2m +1) }{(m+1)} \le \frac{1}{\sqrt{3m+4}} \quad \quad ......(ii)$$

Now ,
$$12m>3m$$ $$12m + 4 > 3m +4$$ $$\sqrt{12m +4} > \sqrt{3m +4}$$ $$\frac{1}{\sqrt{12m +4}} < \frac{1}{\sqrt{3m +4}}$$

Then, I thought that if $\frac{(2m +1) }{(m+1)}$ is smaller than $1$ , i would get inequality (i) but this is false. So I am lost. Can anyone help me to proceed further to get the desired result?

Thanking in advance

Vedant Chhapariya
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  • In the original assertion, in the LHS numerator, does $2n!$ represent $2 \times (n!)$ or does it represent $(2n)! ~$ ? Based on your analysis, it looks like you are interpreting the problem to represent the latter, rather than the former. In your posting, please edit the original assertion to clarify what is intended. – user2661923 Apr 11 '22 at 12:56

1 Answers1

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I think you messed something up in the computation.

If you want to prove this using induction, what you need to show is that \begin{equation*} \frac{1}{\sqrt{3n+1}}\frac{(2n+1)(2n+2)}{4(n+1)^2}\leq \frac{1}{\sqrt{3n+4}}. \end{equation*}

After simplifying this means \begin{equation*} \frac{(2n+1)^2}{(2n+2)^2}\leq \frac{3n+1}{3n+4}, \end{equation*} which you can easily check by direct computation.

AnCar
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