Textbook Problem

Let $I_n={\displaystyle {\int }_0^{\pi /2}{\sin }^{n}xdx}.$

(a) Show that $I_{2n+2}\le I_{2n+1}\le I_{2n}$.

(b) Use Exercise 50 to show that

$\frac{I_{2n+2}}{I_{2n}}=\frac{2n+1}{2n+2}$

(c) Use parts (a) and (b) to show that

$\frac{2n+1}{2n+2}\le \frac{I_{2n+1}}{I_{2n}}\le 1$

and deduce that ${\lim }_{n\to \infty }{I}_{2n+1}/I_{2n}=1.$

(d) Use part (c) and Exercises 49 and 50 to show that

$\underset{n\to \infty }{\lim }\frac{2}{1}\cdot \frac{2}{3}\cdot \frac{4}{3}\cdot \frac{4}{5}\cdot \frac{6}{5}\cdot \frac{6}{7}\cdot \cdot \cdot \cdot \cdot \frac{2n}{2n-1}\cdot \frac{2n}{2n+1}=\frac{\pi }{2}$

This formula is usually written as an infinite product:

$\frac{\pi }{2}=\frac{2}{1}\cdot \frac{2}{3}\cdot \frac{4}{3}\cdot \frac{4}{5}\cdot \frac{6}{5}\cdot \frac{6}{7}\cdot \cdot \cdot \cdot$

and is called the Wallis product.

(e) 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.