What is (1-1/4)(1-1/9)(1-1/16)....(1-1/n^2)? - via Mind your decision 2 Responses to Find the product superman had this to say about that: October 14, 2014 at 12:41 am 1-1/n^2 = (n^2-1)/n^2 = (n-1)(n+1)/n^2 (1-1/(n-1)^2)(1-1/n^2) = (n-2)(n)/(n-1)^2 * (n-1)(n+1)/n^2 = (n-2)(n+1)/((n-1)n) Required answer = 1*(n+1)/(2*n) Dinesh had this to say about that: October 14, 2014 at 2:10 am The general term is \{ (1-\frac{1}{n^2}) \} which is the same as \{ \frac{(n-1)(n+1)}{n^2} \}. Separating into two products, we get which after telescopic cancellation gives \{ \frac{N+1}{2N} \} Subscribe to comments

1-1/n^2 = (n^2-1)/n^2 = (n-1)(n+1)/n^2

(1-1/(n-1)^2)(1-1/n^2) = (n-2)(n)/(n-1)^2 * (n-1)(n+1)/n^2 = (n-2)(n+1)/((n-1)n)

Required answer = 1*(n+1)/(2*n)

The general term is \{ (1-\frac{1}{n^2}) \} which is the same as \{ \frac{(n-1)(n+1)}{n^2} \}. Separating into two products, we get

which after telescopic cancellation gives \{ \frac{N+1}{2N} \}