Let I n = ∫ 0 π / 2 sin n x   d x .

(a) Show that I 2 n + 2 ≤ I 2 n + 1 ≤ I 2 n .

(b) Use Exercise 50 to show that

I 2 n + 2 I 2 n = 2 n + 1 2 n + 2

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

2 n + 1 2 n + 2 ≤ I 2 n + 1 I 2 n ≤ 1

and deduce that lim n → ∞ I 2 n + 1 / I 2 n = 1.

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

lim n → ∞ 2 1 ⋅ 2 3 ⋅ 4 3 ⋅ 4 5 ⋅ 6 5 ⋅ 6 7 ⋅ ⋅ ⋅ ⋅ ⋅ 2 n 2 n − 1 ⋅ 2 n 2 n + 1 = π 2

This formula is usually written as an infinite product:

π 2 = 2 1 ⋅ 2 3 ⋅ 4 3 ⋅ 4 5 ⋅ 6 5 ⋅ 6 7 ⋅ ⋅ ⋅ ⋅

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.