An algebra problem by Ajala Singh

1-1/2+1/4-1/8...

is a geometric series with a common ratio of -1/2. The formula for the sum of a geometric series is:

$\sum_{i=1}^{\infty} (-1/2)^{n-1} = \frac{1-{-1/2}^\infty}{1-(-1/2)} = \frac{1-0}{1+1/2} = \frac{1}{3/2} = 2/3$

see...above prblm can be written as.. (1+1/4+1/16+...)- (1/2+1/8+1/32...) =(USING G.P PROGRESSION...Sn = a/(1-r) valid..only for those sums..in which value extends to 1/∞ as in above expression..) (here a is 1st term and r is the ratio..between two consecutive number) 4/3-2/3=2/3