I can't see how symmetric functions would help. I'm looking for a closed form solution for the given sum, in the sense that the infinite sum of a*r^n = a*(1-r^(n+1))/(1-r), I'm looking for the infinite sum of a*r^(1/n).
Could you write what you mean, rather than abbreviating it? I can't tell precisely what you mean, and my best guesses for what you mean are very obviously not convergent sums.
What does \sum_{k=0}^{n} a*r^(1/k) equal? Given that |r| < 1, a and r are constants.
In the sense that the geometric progression \sum_{k=0}^{n} a*r^k equals a*(1-r^(n+1))/(1-r).
The n'th roots of any real number, say r, is [itex]r^{\frac{1}{n}} \zeta_n^k[/itex] where [itex]\zeta_n[/itex] is the primitive nth root of unity. So what will happen when you sum them?