The uniform convergence of a power series establishes that the function it represents must be continuous. Not only that, but it turns out that the limiting function must be differentiable.
A side note here: we define the derivative of a complex function by exactly the same limit of a difference quotient as before. There’s a lot to be said about derivatives of complex functions, but we’ll set the rest aside until later.
Now, to be specific: if the power series converges for to a function , then has a derivative , which itself has a power series expansion
which converges within the same radius .
Given a point within of , we can expand as a power series about :
convergent within some radius of . Then for in this smaller disk of convergence we have
by manipulations we know to work for series. Then the series on the right must converge to a continuous function, and continuity tells us that each term vanishes as approaches . Thus exists and equals . But our formula for tells us
Finally, we can apply the root test again. The terms are now . Since the first radical expression goes to , the limit superior is the same as in the original series for : . Thus the derived series has the same radius of convergence.
Notice now that we can apply the exact same reasoning to , and find that it has a derivative , which has a power series expansion
which again converges within the same radius. And so on, we determine that the limiting function of the power series has derivatives of arbitrarily large orders.