Let $I_n = \int_{0}^{\,\pi/2} {\sin^n(x)}\,dx.$

One can show that $$I_{2n+2} \le I_{2n+1} \le I_{2n}$$ and $$\frac{I_{2n+2}}{I_{2n}} = \frac{2n+1}{2n+2}.$$ These two facts can then be used to show that $$\frac{2n+1}{2n+2} \le \frac{I_{2n+1}}{I_{2n}} \le 1$$ and deduce that $$\lim_{n \to \infty} \frac{I_{2n+1}}{I_{2n}} = 1.$$ $$\lim_{n \to \infty} \frac{2}{1} \cdot \frac{2}{3} \cdot \frac{4}{3} \cdot \frac{4}{5} \cdot \frac{6}{5} \cdot \frac{6}{7} \cdot \cdots \cdot \frac{2n}{2n-1} \cdot \frac{2n}{2n+1} = \frac{\pi}{2}.$$ This formula is usually written as an infinite product $$\frac{\pi}{2} = \frac{2}{1} \cdot \frac{2}{3} \cdot \frac{4}{3} \cdot \frac{4}{5} \cdot \frac{6}{5} \cdot \frac{6}{7} \cdot \cdots$$ and is called the Wallis product . You should work through the above description to make sure that you understand what is going on!

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 limits of ratios of the width to the height of these rectangles.

Limiting ratio of width to height is