Hi!The series you have described is not a geometric series. It is an example of a more general class of series called power series, which are of the form
My question is about geometric series. I read about the one that you solved, but this one is a little bit different :
What is the sum from i = 0 to infinity of (x^i)(i^2)?
inf - n \ a x / n - n=0where the coefficients a_n don't depend on the variable x. In your example, a_n = n^2.
A key fact about power series is that, if the series converges on an interval of the form |x| < R, then it "converges uniformly" on any closed subinterval of that interval. I won't attempt to explain what that means, but will mention instead an important consequence (which is not always true for series that are not power series): the series can be integrated and differentiated term by term, in the sense that, if you define
inf - n f(x) = \ a x , / n - n=0
inf - n-1 f'(x) = \ n a x . / n - n=0
This means that, if you start with the geometric series
inf - n \ x / - n=0which is known to converge to 1/(1-x) when |x| < 1 (as described in the answer to another question), the following is true for all |x| < 1 by differentiating both sides of the equation:
inf 1 - n-1 ----- = \ n x . 2 / (1-x) - n=0
If you multiply both sides by x you get something close to what you want:
inf x - n ----- = \ n x . 2 / (1-x) - n=0
Differentiating both sides again and multiplying by x again gives you what you want:
inf 1 + x - 2 n x ----- = \ n x . 3 / (1-x) - n=0
Therefore, your series converges to (x+x^2)/(1-x)^3, provided |x| < 1. (If |x| > 1, it diverges).
This particular technique will, of course, work only for this specific example, but the general method for finding a closed-form formula for a power series is to look for a way to obtain it (by differentiation, integration, etc.) from another power series whose sum is already known (such as the geometric series, or a series you can recognize as the Taylor series of a known function).
Most series don't have a closed-form formula, but for those that do, the above general strategy usually helps one to find it.