Math Pre-Calculus 11 (Help)

Find the sum of each infinite series. the question is in the link. https://www.dropbox.com/s/pxz3e2ky9s76hze/IMAG0527.jpg ... the answer in my book is " -4/15 " and please show any specific formulas and what not. thanks.

a1 = -(2/3)^2 = -4/9
r = -2/3
sum = a1/(1-r) = (-4/9)/(1+2/3) = -4/(9+6) = -4/15

{SUMi=1,inf}: ((-1)^i)×(2/3)^(i+1) = [{SUMi=1,inf}: (2/3)^(2i+1)] - [{SUMi=1,inf}: (2/3)^(2i)] =
[(2/3)×{SUMi=1,inf}: (4/9)^i] - [{SUMi=1,inf}: (4/9)^i] =
[(2/3) × (0 - 4/9)/(4/9 - 1)] - (0 - 4/9)/(4/9 - 1) =
(2/3)×(4/5) - (4/5) =
-4/15