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

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.

• There is a proof of #2 without induction here. It is shown here as well. Dec 31, 2016 at 7:17
• @StubbornAtom thanks, what about 1) and 2)? Dec 31, 2016 at 7:20
• Why don't you show what you have tried and exactly where you are stuck? Dec 31, 2016 at 7:23
• Please, post only one question in one post. Posting several questions in the same post is discouraged and such questions may be put on hold, see meta. (OTOH maybe these ones are close enough to each other. It's basically a judgement call.) Dec 31, 2016 at 9:30
• First define $\prod_{k=1}^na_k$ without recursion. Then we can prove statements about it without induction. Dec 31, 2016 at 12:15

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.

• You did 3, I did 1. What a team! Dec 31, 2016 at 7:30
• @martycohen Yeah, and $2$ is already in the comments, time for a well deserved break ! Good job partner. Dec 31, 2016 at 7:32
• thanks to the partners, by the way! Dec 31, 2016 at 7:33
• Nice, but how do you prove that the arithmetic mean of $1,3,\dots,2n-1$ is $n$ without induction? Dec 31, 2016 at 11:02
• And then there is the philosophical point that any statement which has "for all n" implies the use of induction in its proof. Dec 31, 2016 at 15:24

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.

• My edit was a typo jn the first inequality; RHS changed from $1/2^2$ to $1/2^n$ Dec 31, 2016 at 9:56

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 $\frac{1}{2}$.

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}

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}

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}

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)

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

• thanks, by the way, is it somehow related to erdös proof of bertrand's postulate? Dec 31, 2016 at 10:56
• An estimate on $\binom{2n}n$ is used in Erdős' proof of Bertrand's postulate and some other proofs in number theory. However, I am not sure whether this one is used somewhere. Dec 31, 2016 at 11:02