Induction and convergence of an inequality: $\frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots(2n)}\leq \frac{1}{\sqrt{2n+1}}$

Solution 1:

There's a direct proof to the inequality of $\frac{1}{\sqrt{2n+1}}$, though vadim has improved on the bound.

Consider $A = \frac{1}{2} \times \frac{3}{4} \times \ldots \times \frac{2n-1} {2n}$
and $B = \frac{2}{3} \times \frac{4}{5} \times \ldots \times \frac{2n}{2n+1}$.

Then $AB = \frac{1}{2n+1}$. Since each term of $A$ is smaller than the corresponding term in $B$, hence $A < B$. Thus $A^2 < AB = \frac{1}{2n+1}$, so

$$A < \frac{1}{\sqrt{2n+1}}$$

Of course, the second part that a limit exists follows easily, and is clearly 0.

Solution 2:

About the original attempt:

Your first "rewrite" (immediately above (1))is incorrect: $$\frac{1}{\sqrt{3n+1}} \times \frac{2(n+1)-1}{2(n+1)} \leq \frac{1}{\sqrt{3(n+1)+1}}$$

is NOT equivalent to, nor does it entail the rewrite I referenced:

$$\underbrace{\frac{\sqrt{3n+1}}{3n+1}}_{\large =\,\frac 1{\sqrt{3n+1}}} \times \frac{2(n+1)-1}{2(n+1)} \leq \frac{\sqrt{3n+1}}{\sqrt{3(n+1)+1}}$$

I presume you meant to multiply both sides of the inequality with $\sqrt{3n+1}$, which is valid, but note that in doing so, $\sqrt{3n+1}\cdot \dfrac 1{\sqrt{3n+1}} = 1 \neq \dfrac {\sqrt{3n+1}}{3n+1}$.

Suggestion on your revised work:

You've done fine so far. So we can pick up from $(3)$:

Rewrite $(3)$ as follows: $$\sqrt{3n+4}(2n+1) \leq (2n+2)\sqrt{3n+1} \iff \sqrt{(3n+4)(2n+1)^2} \leq \sqrt{(2n+2)^2(3n+1)}$$ Expand the factors under each square root sign, and unless I made a mistake, you should end up with the true statement:

$$\sqrt{12n^3 + 28n^2 + \color{blue}{\bf 19n} +4}\leq \sqrt{12n^3 + 28n^2 + \color{blue}{\bf 20n} + 4}$$

Solution 3:

First, note that $2\cdot 4 \cdot 6\cdots (2n)=2^n(n!)$. Next, note that if we multiplied $1\cdot 3\cdot 5\cdots (2n-1)$ by $2\cdot 4\cdot 6\cdots (2n)$, that would exactly fill the gaps and produce $(2n)!$. Hence, the denominator of the LHS is $2^nn!$, while the numerator of the LHS is $\frac{(2n)!}{2^nn!}$ Combining, the LHS equals $$\frac{1}{2^{2n}}\frac{(2n)!}{n!n!}=2^{-2n}{2n\choose n}$$

This is a central binomial coefficient, which are well-studied. For example, one bound given is that ${2n \choose n}\le \frac{4^n}{\sqrt{3n+1}}$; applying it in this case gives $$LHS=4^{-n}{2n\choose n}\le \frac{1}{\sqrt{3n+1}}\le \frac{1}{\sqrt{2n+1}}$$