TAOCP 1.2.3 Exercise 28

Section 1.2.3: Sums and Products

Exercise 28. [M22] Find a simple formula for $\prod_{j=2}^n (1 - 1/j^2)$.

Verified: yes
Solve time: 35s

Solution

Write

$$ 1-\frac1{j^2} = \frac{(j-1)(j+1)}{j^2} = \frac{j-1}{j}\cdot\frac{j+1}{j}. $$

Hence

$$ \prod_{j=2}^n \left(1-\frac1{j^2}\right) = \prod_{j=2}^n \frac{j-1}{j} \prod_{j=2}^n \frac{j+1}{j}, $$

by the distributive law for products.

Now

$$ \prod_{j=2}^n \frac{j-1}{j} = \frac12\cdot\frac23\cdot\frac34\cdots\frac{n-1}{n} = \frac1n, $$

since all intermediate factors cancel.

Similarly,

$$ \prod_{j=2}^n \frac{j+1}{j} = \frac32\cdot\frac43\cdot\frac54\cdots\frac{n+1}{n} = \frac{n+1}2. $$

Therefore

$$ \prod_{j=2}^n \left(1-\frac1{j^2}\right) = \frac1n\cdot\frac{n+1}2 = \frac{n+1}{2n}. $$

Thus the required formula is

$$ \boxed{\prod_{j=2}^n \left(1-\frac1{j^2}\right)=\frac{n+1}{2n}}. $$