Note that $s_{n+1}=\frac{3}{2}\frac{5}{4}...\frac{2n+2-1}{2n+2-2}=s_n\frac{2n+1}{2n}$ Note that $$ \frac{\pi(n-1)}{2n}<W<\frac{\pi (n+1)}{2n} $$ It is now clear that as $n \rightarrow \infty$, $W \rightarrow \frac{\pi}{2}$ It's worth taking a second look at the way $\pi$ appears in the proof. As we know $\pi$ is deeply interconnected with geometry and it's surprisingly refreshing how the author makes the bridge and by using the circle and its area brings $\pi$ into the game. A partial sum with an even number of factors is for instance $$ \frac{2}{1}\cdot \frac{2}{3} \cdot \frac{4}{3} \cdot \frac{4}{5} $$ which has 4 factors. As we can see for even factors we always end up at the end with a term of the form $$ \frac{2n-2}{2n-1} $$ In the case above $n=3$, $\frac{2\cdot3-2}{2\cdot 3-1}=\frac{4}{5}$. It's not difficult to see that for any even number of factors $$ \frac{2^2\cdot4^2...(2n-2)}{1\cdot3^2...(2n-3)^2}\cdot \frac{2n-2}{2n-1}= \\ = \frac{2^2\cdot4^2...(2n-2)^2}{1\cdot3^2...(2n-3)^2}\cdot \frac{1}{2n-1} = \\ = \frac{2n-1}{s_n^2} $$ John Wallis was an English mathematician who is given partial credit for the development of infinitesimal calculus. He is also credited with introducing the symbol $\infty$ for infinity. He similarly used $1/\infty$ for an infinitesimal.  Wallis made significant contributions to trigonometry, calculus, geometry, and the analysis of infinite series. In his Opera Mathematica I (1695) Wallis introduced the term "continued fraction". If we group all $R_{i,j}$ with the same $i+j$ we get the following "stripes"  The area of each "stripe" is by identity (5) equal to 1 $$ R_{0,0}=1 \\ R_{1,0}+R_{0,1}=1 \\ R_{2,0}+R_{0,2}+R_{1,1}=1 \\ R_{3,0}+R_{0,3}+R_{2,1}+R_{1,2}=1 \\ $$ And so it is easy to conclude that the area of $P_n=n$ Note that $$ \frac{2j+1}{2(j+1)}\frac{j+1}{i+j+1}+\frac{2i+1}{2(i+1)}\frac{i+1}{i+j+1}= \\ = \frac{2j+1}{2(j+1)}\frac{j+1}{i+j+1}\frac{i+1}{i+1}+\frac{2i+1}{2(i+1)}\frac{i+1}{i+j+1}\frac{j+1}{j+1}= \\ = \frac{2(j+1)(i+j+1)(i+1)}{2(j+1)(i+j+1)(i+1)}=1 $$ A partial sum with an odd number of factors is for instance $$ \frac{2}{1}\cdot \frac{2}{3} \cdot \frac{4}{3} \cdot \frac{4}{5} \cdot \frac{6}{5} $$ which has 5 factors. As we can see for odd factors we always end up at the end with a term of the form $$ \frac{2n}{2n-1} $$ In the case above $n=3$, $\frac{2\cdot3}{2\cdot 3-1}=\frac{6}{5}$. It's not difficult to see that for any odd number of factors $$ \frac{2^2\cdot4^2...(2n-2)^2}{1\cdot3^2...(2n-1)}\cdot \frac{2n}{2n-1}= \\ = \frac{2^2\cdot4^2...(2n-2)^2}{1\cdot3^2...(2n-1)^2}\cdot 2n = \\ = \frac{2n}{s_n^2} $$ This is the proof present in most textbooks using integration and recursion. Consider $J_n=\int^{\pi/2}_0 \cos^n (x) dx$. Integrating by parts with $u=\cos^{n-1}(x)$ and $dv=\cos(x)$ we have $$ \int_0^{\pi/2}\cos^n (x) dx = (n-1)\int_0^{\pi/2} \cos^{n-2} (x)\sin^2 (x) dx = \\ = (n-1)\int_0^{\pi/2} \cos^{n-2} (x)(1-\cos^2 (x)) dx= \\ = (n-1)\int_0^{\pi/2} \cos^{n-2} (x) dx - (n-1)\int_0^{\pi/2} \cos^{n} (x) dx $$ Reorganizing the terms we have $$ nJ_n=(n-1)J_{n-2} $$ Now $J_1=1$ and $J_3=2/3$, $J_5=\frac{2\cdot4}{3\cdot5}$ which means that $$ J_{2n+1}=\frac{2\cdot4...(2n-2)\cdot(2n)}{1\cdot 3 ... (2n-1)(2n+1)} $$ At the same time $J_2=\frac{\pi}{2\cdot 2}$, $J_4=\frac{3 \cdot \pi}{2\cdot 4 \cdot 2}$ and $J_6=\frac{3\cdot 5 \cdot \pi}{2\cdot 4\cdot 6\cdot 2}$ which means that $$ J_{2n}=\frac{3\cdot 5 ... (2n-3) \cdot (2n-1) \cdot \pi }{2 \cdot 4...(2n-2)\cdot (2n) \cdot 2} $$ For $0\leq x \leq \frac{\pi}{2}$, $0\leq \cos (x) \leq 1$ so $\cos^{2n} (x) \geq \cos^{2n+1} (x)\geq \cos^{2n+2} (x)$ which implies that $J_{2n} \geq J_{2n+1}\geq J_{2n+2} $. Finally $$ 1 \geq \frac{J_{2n+1}}{J_{2n}} \geq \frac{J_{2n+2}}{J_{2n}} = \frac{2n+1}{2n+2} $$ This was actually the way that Wallis himself derived the formula. He compared $\int _{0}^{\pi }\sin ^{n}xdx$ for even and odd values of n, and noted that for large $n$, increasing $n$ by 1 results in a change that becomes ever smaller as $n$ increases. Since infinitesimal calculus as we know it did not yet exist then, and the mathematical analysis of the time was inadequate to discuss the convergence issues, this was a hard piece of research. Wallis' product is, in retrospect, an easy corollary of the later Euler formula for the sine function: $$ \frac{\sin x}{x} = \prod_{n=1}^\infty\left(1 - \frac{x^2}{n^2\pi^2}\right) $$ Let $x = \frac{π}{2}$ $$ \frac{2}{\pi} = \prod_{n=1}^{\infty} \left(1 - \frac{1}{4n^2}\right) \\ \frac{\pi}{2} = \prod_{n=1}^{\infty} \left(\frac{4n^2}{4n^2 - 1}\right) = \\ = \prod_{n=1}^{\infty} \left(\frac{2n}{2n-1}\cdot\frac{2n}{2n+1}\right) = \frac{2}{1} \cdot \frac{2}{3} \cdot \frac{4}{3} \cdot \frac{4}{5} \cdot \frac{6}{5} \cdot \frac{6}{7} \cdots $$ In 2015 researchers in a surprise discovery, found the same formula in [quantum mechanical calculations of the energy levels of a hydrogen atom](http://phys.org/news/2015-11-derivation-pi-links-quantum-physics.html).