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

Question

Problem statement: Prove that $\frac{1*3*5*...*(2n-1)}{2*4*6*...(2n)}\leq \frac{1}{\sqrt{2n+1}}$ and that there exists a limit when $n \to \infty $. , $n\in \mathbb{N}$

My progress

LHS is equivalent to $\frac{(2n-1)!}{(2n)!}=\frac{(2n-1)(2n-2)(2n-3)\cdot ....}{(2n)(2n-1)(2n-2)\cdot ....}=\frac{1}{2n}$ So we can rewrite our inequality as:

$\frac{1}{2n}\leq \frac{1}{\sqrt{2n+1}}$ Let's use induction:

For $n=1$ it is obviously true. Assume $n=k$ is correct and show that $n=k+1$ holds.

$\frac{1}{2k+2}\leq \frac{1}{\sqrt{2k+3}}\Leftrightarrow 2k+2\geq\sqrt{2k+3}\Leftrightarrow 4(k+\frac{3}{4})^2-\frac{5}{4}$ after squaring and completing the square. And this does not hold for all $n$

About convergence: Is it not enough to check that $\lim_{n \to \infty}\frac{1}{2n}=\infty$ and conclude that it does not converge?

Answer 1

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 $$LHS=4^{-n}{2n\choose n}\le \frac{1}{\sqrt{3n+1}}\le \frac{1}{\sqrt{2n+1}}$$

Answer 2

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

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

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

Answer 3

It boils down to showing

$$\frac{2k+1}{2k+2} \leqslant \frac{\sqrt{3k+1}}{\sqrt{3k+4}}$$

for $k \geqslant 2$. Squaring that, we need to show

$$\left(1-\frac{1}{2k+2}\right)^2 = 1 - \frac{1}{k+1} + \frac{1}{4(k+1)^2} \leqslant 1 - \frac{3}{3k+4}.$$

From then on, it should not be difficult.

Answer 4

Suppose that $P(n)$ which is $\frac{1}{2}\frac{3}{4} \dots \frac{2n-1}{2n} \lt \frac{1}{\sqrt{3n+1}}$. Now $P(n + 1)$ would be $\frac{1}{2}\frac{3}{4} \dots \frac{2n-1}{2n}\frac{2n+1}{2n+2} \overset{?}{\lt} \frac{1}{\sqrt{3n+4}} $. Using the induction hypothesis it would be enough to prove that $\frac{1}{\sqrt{3n+1}}\frac{2n+1}{2n+2} \lt \frac{1}{\sqrt{3n+4}} \leftrightarrow (2n+1)^2(3n+4) \lt (2n+2)^2(3n+1) \leftrightarrow 12n^3+28n^2+19n+4 $$\lt 12n^3+28n^2+20n+4$ which is true.

Answer 5

It is enough to show that $$\frac{1}{\sqrt{3k+1}} \cdot \frac{2k+1}{2k+2} <\frac{1}{\sqrt{3k+4}}$$ or equivalently $$ (2k+1)^2 (3k+4) <(2k+2)^2 (3k+1) $$ but the last is easy to check out.

Answer 6

Suppose $$\frac{2(k+1)-1}{2(k+1)}.\frac{1}{\sqrt{3k+1}}\geq \frac{1}{\sqrt{3(k+1)+1}}$$

$$ \frac{2k+1}{2(k+1)}.\frac{1}{\sqrt{3k+1}}\geq \frac{1}{\sqrt{3(k+1)+1}}$$

$$\frac{(2k+1)^2}{4(k+1)^2 (3k+1)}\geq \frac{1}{3(k+1)+1}$$

$$ 3(k+1)(2k+1)^2+(2k+1)^2 \geq 4(k^2+2k+1)(3k+1)$$

$$ (3k+3)(4k^2+4k+1)+4k^2+4k+1 \geq (k^2+2k+1)(12k+4)$$

$$12k^3+12k^2+12k^2+12k+3k+3+4k^2+4k+1 \geq 12k^3+24k^2+12k+4k^2+8k+4$$

$$19k \geq 20k$$

$$ 1\geq k$$

But .....

Answer 7

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, $\sqrt{3n+1}\cdot \dfrac 1{\sqrt{3n+1}} = 1 \neq \dfrac {\sqrt{3n+1}}{3n+1}$.

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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:

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