>>11597784Proving that they have to be equal whenever the series converges is doable many ways, since there are all sorts of ways to get equality of formal series (convergence sold separately). e.g. the easiest one being that multiplying out (1-x)(1+x+x^2...) gives you 1, so you can plug in any value of x where the sum exists and you'll get 1 there too.
Or, the formal derivative of the geometric series satisfies G' = G^2, so the closed form has to be the solution of the IVP f^2 = f', f(0) = 1, which is 1/(1-x).
You could also use Newton's binomial theorem:
Or, as brought up already, you can build a Taylor series.
Then you need to prove it converges on (-1,1) in the first place, but there are at least a few other ways to do this too, I think. You can chain together a comparison test and the p-series theorem, since if |a| < 1 then eventually |a|^p < (1/p)^2. Integral test would work too, since for |a| < 1.
This is all kind of dumb though. Just do it the normal way.
