Question: Find the sum: 1 + 2x + 3x square + 4x cube +...............+ infinity.
We can express the sum as a sum of sums.
s1 = 1 + x + x^2 + x^3 + x^4 + x^5 + ...
s2 = x + x^2 + x^3 + x^4 + x^5 + ...
s3 = x^2 + x^3 + x^4 + x^5 + ...
s4 = x^3 + x^4 + x^5 + ...
s5 = x^4 + x^5 + ...
s6 = x^5 + ...
etc.
Summing the above sums gives us S = s1 + s2 + s3 + s4 + s5 + s6 + ... , i.e.
S = 1 + 2x + 3x^2 + 4x^3 + 5x^4 + 6x^5 + ...
s1 = sum(k = 1.. infty] x^(k-1) = 1/(1-x), x < 1
s2 = sum(k = 2.. infty] x^(k-1) = x/(1-x), x < 1
s3 = sum(k = 3.. infty] x^(k-1) = x^2/(1-x), x < 1
s4 = sum(k = 4.. infty] x^(k-1) = x^3/(1-x), x < 1
s5 = sum(k = 5.. infty] x^(k-1) = x^4/(1-x), x < 1
s6 = sum(k = 6.. infty] x^(k-1) = x^5/(1-x), x < 1
etc.
So now,
S = s1 + s2 + s3 + ... = 1/(1-x){1 + x + x^2 + x^3 + ...} = {1/(1-x)}{1/(1-x)} = 1/(1 - x)^2
Answer: Sum = 1/(1 - x)^2, x < 1