4
$\begingroup$

Prove for all $n$: $\frac{1}{2}\cdot\frac{3}{4}\cdot...\cdot\frac{2n-1}{2n}<\frac{1}{\sqrt{3n}}$.

Using induction, I tried the brain-dead method and went straight for $$\frac{2n+1}{2n+2}\cdot\frac{1}{\sqrt{3n}}<\frac{1}{\sqrt{3n+3}}$$ $$...$$ $$1<0.$$ After embarrassing myself, I looked around and I found this thread. Using induction, we can then easily prove $$\frac{1}{2}\cdot\frac{3}{4}\cdot...\cdot\frac{2n-1}{2n}\leq\frac{1}{\sqrt{3n+1}}$$ $$\frac{1}{2}\cdot\frac{3}{4}\cdot...\cdot\frac{2n-1}{2n}\leq\frac{1}{\sqrt{3n+1}}<\frac{1}{\sqrt{3n}}.$$ This gets me to the original problem. But in a problem solving standpoint, how do you think to use $\frac{1}{\sqrt{3n+1}}$? Is there some point in the first induction that leads to this idea? Or is there a better method than the above?

$\endgroup$
1
  • 2
    $\begingroup$ You might find this thread interesting. In fact, this exact question shows up as an answer, but no explanation is given. $\endgroup$ Commented Dec 18, 2020 at 6:53

3 Answers 3

3
$\begingroup$

Write $a_n$ for the $n$th term in your sequence. Look at the square of $a_n$, and rotate the numerators left by one position. Starting at $n=2$ you observe $$ a_2^2=\frac12\frac12\frac34\frac34=\frac12\left(\frac32\frac34\right)\frac14\\ a_3^2=\frac12\frac12\frac34\frac34\frac56\frac56=\frac12\left(\frac32\frac34\right)\left(\frac54\frac56\right)\frac16\\ a_4^2=\frac12\frac12\frac34\frac34\frac56\frac56\frac78\frac78=\frac12\left(\frac32\frac34\right)\left(\frac54\frac56\right)\left(\frac76\frac78\right)\frac18 $$ and so on. The inequality $1+x\le e^x$ then gives $$ a_2^2\le \frac18 \exp\left(\frac18\right)\\ a_3^2\le\frac1{12}\exp\left(\frac18+\frac1{24}\right)\\ a_4^2\le\frac1{16}\exp\left(\frac18+\frac1{24}+\frac1{48}\right) $$ and in general for $n\ge 2$ $$a_n^2\le \frac1{4n}\exp\left[\frac14\left(\frac12+\frac16+\frac1{12}+\cdots+\frac1{n(n-1)}\right)\right].$$ The series $\frac12+\frac16+\frac1{12}+\cdots+\frac1{n(n-1)}$ telescopes to $1$, yielding $$a_n^2\le \frac{e^{1/4}}{4n}$$ which also holds for $n=1$. Since $e^{1/4}\approx 1.284 < 4/3$, this proves $a_n^2< \frac1{3n}$.

$\endgroup$
2
$\begingroup$

Notice that $$\frac{1}{\sqrt{an+b}} \cdot \frac{2n+1}{2n+2} \le \frac{1}{\sqrt{a(n+1)+b}} \\ \iff (a(n+1)+b)(2n+1)^2 \le (2n+2)^2 (an+b) \\ \iff an+a-4bn-3b \le 0$$

Hence if $a=3$, then $b=1$ would work. Of course, you need to prove the initial case ($n$=1).

BTW: how amazing it is that the first two answers got $e$ and $\pi$, respectively.

$\endgroup$
1
  • 1
    $\begingroup$ (+1) So as long as $\min(4b-a,3b-a)\ge0$, the induction proceeds. I like this approach (I have used a similar technique a few times). I have added a more elementary approach, based on bounds for $\frac1{4^n}\binom{2n}{n}$ from an earlier answer. $\endgroup$
    – robjohn
    Commented Dec 20, 2020 at 2:49
2
$\begingroup$

A First Approach $$ \begin{align} n\prod_{k=1}^n\left(\frac{2k-1}{2k}\right)^2 &=\frac14\prod_{k=2}^n\left(\frac{2k-1}{2k}\right)^2\frac{k}{k-1}\tag{1a}\\ &=\frac14\prod_{k=2}^n\frac{2k-1}{2k}\frac{2k-1}{2k-2}\tag{1b}\\ &=\frac14\prod_{k=2}^n\frac{\color{#C00}{k-1/2}}{\color{#090}{k}}\frac{\color{#75F}{k-1/2}}{\color{#C90}{k-1}}\tag{1c}\\ &=\frac14\color{#C00}{\frac{\Gamma(n+1/2)}{\Gamma(3/2)}}\color{#090}{\frac{\Gamma(2)}{\Gamma(n+1)}}\color{#75F}{\frac{\Gamma(n+1/2)}{\Gamma(3/2)}}\color{#C90}{\frac{\Gamma(1)}{\Gamma(n)}}\tag{1d}\\[3pt] &=\frac1\pi\frac{\Gamma(n+1/2)^2}{\Gamma(n+1)\,\Gamma(n)}\tag{1e}\\[3pt] &\le\frac1\pi\tag{1f} \end{align} $$ Explanation:
$\text{(1a)}$: pull the $k=1$ term out front and bring $n$ inside as a telescoping product
$\text{(1b)}$: rearrange terms
$\text{(1c)}$: divide numerator and denominator by $2$
$\text{(1d)}$: write the products as ratios of the Gamma function, using $\Gamma(x+1)=x\,\Gamma(x)$
$\text{(1e)}$: collect terms using $\Gamma(1)=\Gamma(2)=1$ and $\Gamma(3/2)=\sqrt\pi/2$
$\text{(1f)}$: $\Gamma(x)$ is log-convex

Thus, we get the stronger $$ \prod_{k=1}^n\frac{2k-1}{2k}\le\frac1{\sqrt{\pi n}}\tag2 $$


A Slightly Simpler Approach with a Better Bound $$ \begin{align} \prod_{k=1}^n\frac{2k-1}{2k} &=\prod_{k=1}^n\frac{(2k-1)2k}{4k^2}\tag{3a}\\ &=\frac1{4^n}\binom{2n}{n}\tag{3b}\\ &\le\frac1{\sqrt{\pi\!\left(n+\frac14\right)}}\tag{3c} \end{align} $$ Explanation:
$\text{(3a)}$: multiply numerator and denominator by $2k$
$\text{(3b)}$: $\prod\limits_{k=1}^n(2k-1)2k=(2n)!$ and $\prod\limits_{k=1}^n2k=2^nn!$
$\text{(3c)}$: inequality $(9)$ from this answer

In fact, using inequality $(10)$ from this answer, we get $$ \frac1{\sqrt{\pi\!\left(n+\frac13\right)}}\le\prod_{k=1}^n\frac{2k-1}{2k}\le\frac1{\sqrt{\pi\!\left(n+\frac14\right)}}\tag4 $$

$\endgroup$
1
  • $\begingroup$ (+1) Cool! Like both approaches. $\endgroup$
    – Neat Math
    Commented Dec 20, 2020 at 2:56

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .