Evaluate the product (1-1/4)(1-1/9)...(1-1/100)

We notice that the factors 1/4,1/9,...1/100 could be
written as:

1/2^2,
1/3^2,...,1/10^2

For calculating the product, we notice
that in the brackets we have differences of squares, type a^2 – b^2=
(a-b)(a+b)

So, (1-
1/2^2)=(1-1/2)(1+1/2)

(1- 1/3^2)=(1-1/3)(1+1/3) and so on
till

(1-1/10^2)=(1-1/10)(1+1/10)

Now,
we’ll do the arithmetical operations in each
paranthesis

(1-1/2)(1+1/2)=(1/2)*(3/2)

(1-1/3)(1+1/3)=(2/3)*(4/3)

…………………………….

(1-1/10)(1+1/10)=(9/10)*(11/10)

We’ll
re-group the factors in a convenient
way:

[(1/2)*(3/2)]*[(2/3)*(4/3)]*….*[
(9/10)*(11/10)]=

=(1/2)*[ (3/2)*(2/3)]*…*[(10/9)*(
9/10)*(11/10)=

P=(1/2)*(11/10)=11/20