Wallis Product and Rectangles

Question

The Question

We construct rectangles as follows. Start with a square of area 1 and attach rectangles of area 1 alternately beside or on top of the previous rectangle (see the figure). Find the limit of the ratios of width to height of these rectangles. (See page 502 exercise 72 of this link for more context.)
What I Have Done
I first tried listing out the first few terms of the sequence of ratios: $$1,2,\frac{4}{3},\frac{16}{9},\frac{64}{45},\cdots$$which turns out the be the first few terms of the Wallis Product: $$1,1\cdot\frac{2}{1},1\cdot\frac{2}{1}\cdot\frac{2}{3},1\cdot\frac{2}{1}\cdot\frac{2}{3}\cdot\frac{4}{3},1\cdot\frac{2}{1}\cdot\frac{2}{3}\cdot\frac{4}{3}\cdot\frac{4}{5},\cdots$$ So a reasonable conjecture is that $$\lim_{n\to\infty}{\frac{w_n}{h_n}}=\prod_{n=1}^\infty{\frac{4n^2}{4n^2-1}}=\frac{\pi}{2}$$ where $w_n$ is the width of the $(2n-1)$th rectangle and $h_n$ is the height of the $(2n-1)$th rectangle. For example $w_1=1,w_2=2$ and $h_1=1,h_2=\frac{3}{2}$. So as our last step, we need to prove it. First I tried constructing recurrence formulas for $w_n$ and $h_n$ then using mathematical induction: \begin{equation} w_n= \begin{cases} 1 & \text{if} & n=1\\\\ w_{n-1}+\frac{1}{h_{n-1}} & \text{if} & n\geq2 \end{cases} \end{equation} \begin{equation} h_n= \begin{cases} 1 & \text{if} & n=1\\\\ h_{n-1}+\frac{1}{1+h_{n-1}w_{n-1}} & \text{if} & n\geq2 \end{cases} \end{equation} Notice that to prove $$\frac{w_n}{h_n}=\prod_{k=1}^{n-1}{\frac{4k^2}{4k^2-1}}$$ for $n\geq2$ (which implies the theorem we want to prove), we need to show that

1. $\frac{w_2}{h_2}=\frac{4\cdot1^2}{4\cdot1^2-1}$ (which is trivial)
2. if $\frac{w_n}{h_n}=\prod_{k=1}^{n-1}{\frac{4k^2}{4k^2-1}}$ for some integer $n\geq2$, it is true for $n+1$

Assume $\frac{w_n}{h_n}=\prod_{k=1}^{n-1}{\frac{4k^2}{4k^2-1}}$ for some integer $n\geq2$, then $$\frac{w_{n+1}}{h_{n+1}}=\prod_{k=1}^{n}{\frac{4k^2}{4k^2-1}}$$ if and only if $$\frac{w_{n+1}/h_{n+1}}{w_n/h_n}=\frac{4n^2}{4n^2-1}$$ and using the recurrence formulas, we get $$\frac{(w_nh_n+1)^2}{w_nh_n(w_nh_n+2)}=\frac{4n^2}{4n^2-1}$$ which I got stuck with. Second I tried searching the Internet. Although I did find an article mentioning it, it did not provide a proof.

Thanks in advance.

Answer 1

Let $u_n$ be the length of the side that was not previously known, i.e. alternatively the height and the width.

We have $u_1=1, u_2=2, u_n=u_{n-2}+1/u_{n-1}$.

Then

• $u_{2n}=\dfrac{4^n}{\binom{2n}{n}}=\dfrac{\pi/2}{W_{2n}}$
• $u_{2n+1}=\dfrac{(2n+1)\binom{2n}{n}}{4^n}=\dfrac{1}{W_{2n+1}}$

where $W_n=\int_0^{\pi/2}\sin^n x\;\mathrm dx$ are Wallis' integrals.

Proof:

It's obvious for $u_1$ and $u_2$.

If the property is true up to $u_{2n}$, then

$$u_{2n+1}=u_{2n-1}+\frac{1}{u_{2n}}=\frac{(2n-1)\binom{2n-2}{n-1}}{4^{n-1}}+\dfrac{\binom{2n}{n}}{4^n}\\ =\frac1{4^n}\left( \frac{4(2n-1)(2n-2)!n^2}{n!^2}+\frac{(2n)!}{n!^2}\right)\\ =\frac1{4^nn!^2}[2n(2n!)+(2n)!]=\frac{(2n+1)(2n)!}{4^nn!^2}$$

Therefore the property is true for $2n+1$.

Likewise, if the property is true up to $2n+1$, then

$$u_{2n+2}=u_{2n}+\frac{1}{u_{2n+1}}=\frac{4^n}{\binom{2n}{n}}+\frac{4^n}{(2n+1)\binom{2n}{n}}\\ =4^n\left( \frac{n!^2}{(2n)!}+\frac{n!^2}{(2n+1)!} \right)\\ =\frac{4^nn!^2}{(2n+2)!}[(2n+2)(2n+1)+(2n+2)]\\ =\frac{4^{n+1}(n+1)!^2}{(2n+2)!}$$

And the property is true for $2n+2$.

Therefore by induction it's true for all $n\ge1$.

The ratio of width to height of the $2n$th rectangle is $\dfrac{u_{2n}}{u_{2n-1}}$, and for the $(2n+1)$th rectangle it's $\dfrac{u_{2n}}{u_{2n+1}}$. And since $W_n\sim W_{n+1}$, both ratios tend to $\frac\pi2$.

Answer 2

When I was bored, I stumbled upon the following answer:
Continuing the "What I've Done" section above, we want to prove that $$\frac{(w_nh_n+1)^2}{w_nh_n(w_nh_n+2)}=\frac{4n^2}{4n^2-1}.$$ Notice that the $(2n-1)$th rectangle has area $(2n-1)$ because every small rectangle has area $1$, that is $$w_nh_n=2n-1.$$ Therefore, \begin{equation*} \begin{split} \frac{(w_nh_n+1)^2}{w_nh_n(w_nh_n+2)}&=\frac{(2n)^2}{(2n-1)(2n+1)}\\ &=\frac{4n^2}{4n^2-1}. \end{split} \end{equation*}