[malisper]

This is pretty neat! I was toying around with the problem and it appears you can use generating functions to derive the same sequence of operations. If you start with:

  G(x) = 1 + x + x^2 + ... = 1/(1-x)

The coefficients of this polynomial is the sequence (0^0, 1^0, 2^0, ...)

If you take the derivative of G(x) and multiply by x you get:

  x * G'(x) = x + 2*x^2 + 3*x^3 + ... = x * d/dx 1/(1-x) = x/(1-x)^2

The coefficients of this polynomial is the sequence (0^1, 1^1, 2^1, ...). If you repeat this step, you get a polynomial whose coefficients are (0^2, 1^2, 2^2, ...) and if you do this operation N times, you can get a closed form of a polynomial whose coefficients are (0^N, 1^N, 2^N, ...).

The infinite sum converges for -1 < x < 1. If you set x=1/c, you get the infinite sum

  0^N/c^0 + 1^N/c^1 + 2^N/c^2 + ...

which is exactly the sum we are trying to solve for. This means you solve any infinite sum of the form given by taking the derivative of 1/(1-x) N times while multiplying by x each time. Then plug in x=1/c at the end.

[ajkjk]

Doing it with generating functions is essentially the same as what is done in the post under a different name.

[shiandow]

True, but the generating functions make it easier to prove that this works, rather than relying on the properties of a particular function (properties that you can most easily prove by reverting to the generating functions).