Xét theo thừa số tổng quát:
\(1-\dfrac{1}{t^2}=\dfrac{t^2-1}{t^2}=\dfrac{\left(t-1\right)\left(t+1\right)}{t^2}\)
Thay vào bài toán:
\(A=\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)\left(1-\dfrac{1}{4^2}\right).....\left(1-\dfrac{1}{n^2}\right)=\left(\dfrac{\left(2-1\right)\left(2+1\right)}{2^2}\right)\left(\dfrac{\left(3-1\right)\left(3+1\right)}{3^2}\right)\left(\dfrac{\left(4-1\right)\left(4+1\right)}{4^2}\right)....\left(\dfrac{\left(n-1\right)\left(n+1\right)}{n^2}\right)\)
\(A=\dfrac{1.3}{2.2}.\dfrac{2.4}{3.3}.\dfrac{3.5}{4.4}......\dfrac{\left(n-1\right)\left(n+1\right)}{n.n}=\dfrac{1.2.3.....\left(n-1\right)}{2.3.4.....n}.\dfrac{3.4.5......\left(n+1\right)}{2.3.4......n}=\dfrac{1}{n}.\dfrac{n+1}{2}=\dfrac{n+1}{2n}\)
