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IIT JEE (1987) asked for the Induction proof of $$P=\frac{(2n)!}{2^{2n} n!^2} \le \frac{1}{\sqrt{3n+1}}~~~~~~(*)$$ which is commonly available. Upon simplification, we get $$P=\frac{1}{2}. \frac{3}{4}. \frac{5}{6} \frac{7}{8}\dots \frac{2n-1}{2n}.$$ In an attempt to produce a non-induction proof, I use $\frac{a}{b}< \frac{a+x}{b+x}$, if $0<a<b$ and $x > 0.$ So, $P<Q$, if $$Q=\frac{2}{3}.\frac{4}{5}.\frac{6}{7}. \dots \frac{2n}{2n+1}.$$ $$ P^2 \le PQ=\frac{1}{2n+1}.$$ $$\implies P< \frac{1}{\sqrt{2n+1}},$$ which is a weaker result than $(*)$.

The question is: What could be a non induction proof of $(*)$?

Z Ahmed
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1 Answers1

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We can prove a stronger version of the previous lemma, which is: $$ P \leq Q \times \frac{2n+1}{3n+1}. $$ I will use the inequality $\frac{Q'}{P} \geq 1$ for the proof, where $Q' = Q \times \frac{2n+1}{3n+1}$.

We have: $$ \frac{Q'}{P} = \frac{4}{3} \times \frac{16}{15} \times \frac{36}{35} \times \dots \times \frac{4n^2}{4n^2-1} \times \frac{2n+1}{3n+1} = f(n), $$ where $$ f(n) = \prod_{k=1}^{n} \frac{4k^2}{4k^2-1} \times \frac{2n+1}{3n+1}. $$ Note that $f(0) = 1$. If we can prove that $f(n)$ is an increasing function in $n$, then the result follows. To do so, we will use the technique $\frac{f(n+1)}{f(n)} \geq 1$.

Now, $$ \frac{f(n+1)}{f(n)} = \frac{4(n+1)^2}{4(n+1)^2-1} \times \frac{(2n+3)(3n+1)}{(3n+4)(2n+1)} \geq 1. $$ Expanding the inequality, we get: $$ (4n^2 + 8n + 4)(6n^2 + 11n + 3) \geq (4n^2 + 8n + 3)(6n^2 + 11n + 4), $$ which simplifies to: $$ 2n^2 + 3n \geq 0. $$ This inequality holds for all $n \geq 0$.

Thus, we have shown that: $$ P^2 \leq P \times Q \times \frac{2n+1}{3n+1} \Rightarrow P^2 \leq \frac{1}{3n+1} \Rightarrow P \leq \frac{1}{\sqrt{3n+1}}. $$

Apologies for the somewhat messy proof.

Arya J
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