For every n ∊ N, (1-(1/2))(1-(1/4))(1-(1/8))(1-(1/16))...(1-(1/2)) is greater than or equal to (1/4) +(1/(2^(n+1)))

 For every n ∊ N, 

(1-(1/2))(1-(1/4))(1-(1/8))(1-(1/16))...(1-(1/2^(n))) is greater than or equal to (1/4) +(1/(2^(n+1)))

PROOF:

Base Case:

For n = 1,

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

1/4 - (1/(2^(n+1))) = 1/4 +(1/2^(1+1)) = 1/4 +1/4 = 2/4 = 1/2

1/2 >= 1/2

Thus the summation holds for n = 1.

Induction Step:

Assume P(n) is true

(1-(1/2))(1-(1/4))(1-(1/8))(1-(1/16))...(1-(1/2^(n))) >= (1/4) +(1/(2^(n+1)))

(1-(1/2))(1-(1/4))(1-(1/8))(1-(1/16))...(1-(1/2^(n)))(1-(1/2^(n+1))) >= (1/4 + (1/2^(n+1)))(1/4 + (1/2^(n+1)))

= 1/4 - (1/2^(n+3)) + (1/2^(n+1)) - (1/2^(2(n+1)))

= 1/4 + (1/2^(n+1))[1- (1/2^(n+1)) -1/4]

*since (1/2^(n+1)) + 1/4 < 1/2

--> 1 - ((1/2^(n+1)) - 1/4) > 1 - 1/2

--> 1 - ((1/2^(n+1)) - 1/4) > 1/2*

> 1/4 +(1/2^(n+1))(1/2) (from the * above)

= 1/4 + (1/2^(n+2)) = (1/2^((n+1)+1))

(1-(1/2))(1-(1/4))(1-(1/8))(1-(1/16))...(1-(1/2^(n+1))) >= 1/4 + (1/2^((n+1)+1))

This proves that P(n+1) is true.

Thus by proof by mathematical induction,  For every n ∊ N, 

(1-(1/2))(1-(1/4))(1-(1/8))(1-(1/16))...(1-(1/2^(n))) is greater than or equal to (1/4) +(1/(2^(n+1)))