Question

Prove that $(1-\frac{1}{2^2})(1-\frac{1}{3^2})(1-\frac{1}{4^2})......(1-\frac{1}{n^2})=\frac{n+1}{2n}$ for all natural numbers, $n\ge 2$.

Answer 1

$(1-\frac{1}{2^2})(1-\frac{1}{3^2})(1-\frac{1}{4^2})......(1-\frac{1}{n^2})$

Above can be written as

$=(-\frac{2^2-1}{2^2})\left(\frac{3^2-1}{3^2}\right)\left(\frac{4^2-1}{4^2}\right)\left(\frac{n^2-1}{n^2}\right)$

$==\left(\frac{(2+1)(4-1)}{4^2}\right)\left(\frac{(3+1)(3-1)}{3^2}\right)\left(\frac{(4+1)(4-1)}{4^2}\right)......\left(\frac{(n+1)(n-1)}{n^2}\right)=\left(\frac{3.1}{2^2}\right)\left(\frac{4.2}{3^2}\right)\left(\frac{5.3}{4^2}\right).......\left(\frac{(n+1)(n-1)}{n^2}\right)$

In the above product, there are two series in numerator 3.4.5 ... (n + 1) and 1.2.3 ...... (n - 1) All numbers from 3 to (n - 1) are repeated twice and 1, 2, n are appeared once in numerator So after cancelling like terms we get

$=\frac{(n+1)}{2n}$