I need to prove that if f is defined for abs(x) < R and if there exists a constant B such that abs(f*(x)) <= B (where f*(x) denotes the n derivatives of f) for all abs(x) < R and n in the natural numbers, then the Taylor Series expansion,

summation from n = 0 to infinity of [f*(0)/(n!)]*x^n converges to f(x) for abs(x) < R.

I did the following work: I have a defined function f(x). Since abs(f*(x)) <= B, it follows that abs(R(x)) <= (abs(x)^n)/n! for n in N and x in R where (R(x)) denotes the sequence of remainders. I know that lim as n goes to infinity of (R(x)) = 0 for each x in R. Because the sequence of remainders converges to 0 for each x in some interval {x: abs(x-c) < R}, I can write that f(x) = summation from n = 0 to infinity of [f*(c)/(n!)]*(x-c)^n. Evaluating at c = 0, the Taylor expansion becomes the summation from n = 0 to infinity of [f*(0)/(n!)]*x^n.

I use the Cauchy-Hadamard Theorem, which states that if R is the radius of convergence of the power series sum(a_n*x^n), then the series is absolutely convergent if abs(x) < R. Since abs(x) < R, I know that the above Taylor series expansion does converge. But how do I show it converges to f(x)?

Thanks in advance for your help.