husoski
(1 - 1/2) (1 + 1/3) (1 - 1/4) (1 + 1/5) (1 - 1/6) .... (1 ± 1/n)
= (1/2) (4/3) (3/4) (6/5) (5/6) .... [(n±1)/n]
Notice the pattern of pairs starting with the 4/3 factor. You have 4/3 followed by 3/4, 6/5 followed by 5/6.
So, with an odd number of factors (meaning n is even), you'll have just 1/2 as the product.
With an even number of factors (meaning n is odd) you'll have the middle n-2 numbers that multiply to 1 and the product will be (1/2)[(n+1)/n] = (n+1)/(2n).
The one thing needed to finish the proof is to show that those odd*even terms always multiply to 1. The product of a odd n term with the next even term is:
[1 + 1/(2k+1)]*[1 - 1/(2k+2)] = [(2k + 2)/(2k + 1)] * [(2k + 1)/(2k + 2)]
The (2k+1)/(2k+1) and (2k+2)/(2k+2) factors cancel out and you're left with:
= 1
QED.