Proof without induction of the inequalities related to product $\prod\limits_{k=1}^n \frac{2k-1}{2k}$

Question

How do you prove the following without induction:

1)$\prod\limits_{k=1}^n\left(\frac{2k-1}{2k}\right)^{\frac{1}{n}}>\frac{1}{2}$

2)$\prod\limits_{k=1}^n \frac{2k-1}{2k}<\frac{1}{\sqrt{2n+1}}$

3)$\prod\limits_{k=1}^n2k-1<n^n$

I think AM-GM-HM inequality is the way, but am unable to proceed. Any ideas. Thanks beforehand.

Answer 1

for problem $3$ notice that the arithmetic average of the numbers $1,3,5,7,\dots,2n-1$ is $n$, and they are also $n$ terms.

So we have $n > \sqrt[n]{1\cdot 3 \cdot 5 \dots (2n-1)}$ by AM-GM.

Raising to the $n$'th power yields the desired result.

Answer 2

1) is equivalent to $\prod\limits_{k=1}^n\frac{2k-1}{2k}>\frac{1}{2^n} $ or $2^n >\prod\limits_{k=1}^n\frac{2k}{2k-1} $.

But $\frac{2k}{2k-1} =1+\frac{1}{2k-1} \le 2 $ for $k \ge 1$ and $\frac{2k}{2k-1} =1+\frac{1}{2k-1} \lt 2 $ for $k \ge 2$ which makes this evident.

Answer 3

Notice that in problem #$1$ if you raise each side to the $n$th power, then it is equivalent to showing that the product of the $n$ factors of the form

$$\left(1-\frac{1}{2k}\right)\tag{1}$$

is greater than $\left(\frac{1}{2}\right)^n$. But that is clearly true since each factor in equation $(1)$ is greater than or equal to $\frac{1}{2}$.

Answer 4

You may notice that the product appearing in 1) and 2) is related to binomial coeffients. $$ \frac{1\cdot3\cdots(2n-1)}{2\cdot4\cdots(2n)}= \frac{(2n)!}{(2\cdot4\cdots(2n))^2}= \frac{(2n)!}{(2^n n!)^2}= \frac1{4^n} \binom{2n}n $$ (See also here.)

So the inequalities in 1) and 2) are in fact equivalent to $$2^n < \binom{2n}n < \frac{4^n}{\sqrt{2n+1}}.$$

There already are several questions on this site about similar (or even stronger) inequalities for the central binomial coefficient.

A sample of such questions (ordered by id)

• Elementary central binomial coefficient estimates
• Prove that $2^n < \binom{2n}{n} < 2^{2n}$
• To prove the inequality:- $\frac{4^m}{2\sqrt{m}}\le\binom{2m}{m}\le\frac{4^m}{\sqrt{2m+1}}$

You can probably find more using internal search, Google or Approach0. Or simply by checking the questions tagged binomial-coefficients+inequality.

Answer 5

1) For $k\ge2$, $$ \frac{2k-1}{2k}\gt\frac12 $$ Therefore, for $n\ge2$ $$ \begin{align} \left[\prod_{k=1}^n\frac{2k-1}{2k}\right]^{1/n} &=\left[\frac12\prod_{k=2}^n\frac{2k-1}{2k}\right]^{1/n}\\ &\gt\left[\frac12\prod_{k=2}^n\frac12\right]^{1/n}\\ &=\left[\prod_{k=1}^n\frac12\right]^{1/n}\\[6pt] &=\frac12 \end{align} $$

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2) Squaring and cross-multiplication show that for $k\ge1$ $$ \frac{2k-1}{2k}\lt\sqrt{\frac{k-\frac12}{k+\frac12}} $$ Therefore, $$ \begin{align} \prod_{k=1}^n\frac{2k-1}{2k} &\lt\prod_{k=1}^n\sqrt{\frac{k-\frac12}{k+\frac12}}\\ &=\sqrt{\frac{\frac12}{n+\frac12}}\\ &=\frac1{\sqrt{2n+1}} \end{align} $$

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3) The AM-GM says $$ \begin{align} \left(\prod_{k=1}^n(2k-1)\right)^{\large\frac1n} &\le\frac1n\sum_{k=1}^n(2k-1)\\[6pt] &=n \end{align} $$