 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2023/discrete-mathematics-probability-and-continuous-mathematics/2/b/solution

n = 2

, the product is

1 - 1/4 = 3/4

, as required. Suppose the formula holds at

n \geq 2

. Appending the next factor gives

\prod_{i = 2}^{n + 1} (1 - \frac{1}{i ^{2}}) = \frac{n + 1}{2 n} \frac{n ( n + 2 )}{( n + 1 ) ^{2}} = \frac{n + 2}{2 ( n + 1 )},

which is exactly the formula at

n + 1

. Thus mathematical induction gives $(n + 1) / (2 n)$ for every

n \geq 2

1 - i^{- 2} = (i - 1) (i + 1) / i^{2}

also explains the cancellation in this telescoping product.

 New to topics? Read the docs here!