Verified answer

(-3)^(n - 1) / 2^(3n) =>

(-3)^n * (-3)^(-1) / (2^3)^n =>

(-3 / 2^3)^n * (-3)^(-1) =>

(-1/3) * (-3/8)^n

(-1/3) * sum((-3/8)^n , n = 1 , n = infinity)

The sum is a geometric sum, and since our common ratio is between -1 and 1, the sum will converge as n goes to infinity. So, let's expand our sum:

S = (-3/8)^1 + (-3/8)^2 + ... + (-3/8)^n

S * (-3/8) = (-3/8)^2 + ... + (-3/8)^n + (-3/8)^(n + 1)

S - S * (-3/8) = (-3/8) + (-3/8)^2 + ... + (-3/8)^n - (-3/8)^2 - (-3/8)^3 - .... - (-3/8)^n - (-3/8)^(n + 1)

S * (1 + (3/8)) = (-3/8) - (-3/8)^(n + 1)

S * (11/8) = (-3/8) - (-3/8)^(n + 1)

As n goes to infinity, (-3/8)^(n + 1) goes to 0

S * (11/8) = (-3/8) - 0

S * (11/8) = -3/8

S = (-3/8) * (8/11)

S = -3/11

(-1/3) * (-3/11) =>

1/11

The sum is 1/11

∑(n= 1 to infinity) [(-3)^(n-1)] / [2^(3n)] = (1/8)∑(n= 0 to infinity) [(-3/8)^n] = (1/8)(1/(1+(3/8)) = 1/11