To prove $\frac{1}{2} \times \frac{3}{4} \times \frac{5}{6} \cdots \frac{(2n-1)}{2n} \le \frac{1}{\sqrt{3n+1}}$

Question

To prove $$P=\frac{1}{2} \times \frac{3}{4} \times \frac{5}{6} \cdots \frac{(2n-1)}{2n}\le \frac{1}{\sqrt{3n+1}}$$

i have written $P$ as

$$P=\frac{(2n)!}{2^{2n}(n!)^2}=\frac{(2n)!}{4^{n}(n!)^2}=\frac{\binom{2n}{n}}{4^n}$$

Now

$$P=\frac{\binom{2n}{n}}{(1+3)^n} \lt \frac{\binom{2n}{n}}{1+3n}$$ since

$$(1+3)^n=1+3n+\binom{n}{2}3^2+\cdots$$

Any help here..

Answer 1

We can use simple induction to find the answer. First, at $n=1$, the inequality holds true. Now let us assume the above for n and prove it for $n+1$.

Let us divide the LHS of the inequality for $n+1$ by LHS of the inequality for n, and do the same with the RHS for $n+1$ and $n$.

$(2n+1)/(2n+2)$ must be lesser than or equal to $\sqrt{(3n+1)}/\sqrt{(3n+3)}$. We can square on both sides.

$(4n^2+4n+1)/(4n^2+8n+4)$ must be lesser than or equal to $(3n+1)/(3n+3)$.

$(4n^2+4n+1)(3n+3)$ must be lesser than or equal to $(4n^2+8n+4)(3n+1)$. This can be easily verified by multiplying the polynomials. Hence, for all '$n$', the inequality holds

Answer 2

We need to prove that $$\binom{2n}{n}\leq\frac{4^n}{\sqrt{3n+1}}.$$ Indeed, by induction for $n=1$ we get an equality.

Now, $$\binom{2n+2}{n+1}=\binom{2n}{n}\cdot\frac{(2n+1)(2n+2)}{(n+1)^2}\leq\frac{4^n}{\sqrt{3n+1}}\cdot\frac{2(2n+1)}{n+1}.$$ Thus, it remains to prove that $$\frac{4^n}{\sqrt{3n+1}}\cdot\frac{2(2n+1)}{n+1}\leq\frac{4^{n+1}}{\sqrt{3n+4}},$$ which is $$(2n+2)^2(3n+1)\geq(2n+1)^2(3n+4)$$ or $$n\geq0.$$

Done!