I guess this is the method that tomarken referred to:

Start with the identity for the sum of a GP..

1 + x + x² + x³ + x⁴ + ..             = 1/(1 – x)

Alternately, differentiate wrt x and multiply by x to regain the powers:

1 + 2x + 3x² + 4x³ +..                = (1–x)^-2

x + 2x² + 3x³ + 4x⁴ +..               = x(1–x)^-2

1 + 2²x + 3²x² + 4²x³ +.. = 2x(1–x)^-3 + (1–x)^-2

x + 2²x² + 3²x³ + 4²x⁴ +..            = 2x²(1–x)^-3 + x(1–x)^-2

1 + 2³x + 3³x² + 4³x³ +.. = 6x²(1–x)^-4 + 4x(1–x)^-3 + 2x(1–x)^-3 + (1–x)^-2

x + 2³x² + 3³x³ + 4³x⁴ +.. = 6x³(1-x)^-4 + 6x²(1-x)^-3 + x(1-x)^-2

Now substituting x = 1/3 gives:

1³/3¹ + 2³/3² + 3³/3³ + 4³/3⁴ + ...   =   33/8

Posted by Harry on 2014-03-06 12:55:19