How do I prove
If there is an $\varepsilon > 0$ such that $\sum_{n=0}^{+\infty} a_nx^n = \sum_{n=0}^{+\infty} b_nx^n$ for $-\varepsilon < x < \varepsilon$, then $a_n = b_n$ for all $n \in \mathbb{N}$.
I know that the $n$-th coefficient in $f(x) = \sum_{n=0}^{+\infty} a_nx^n$ is given by $a_n = \frac{f^{(n)}(0)}{n!}$.
So what I tried: Assume $\sum a_n x^n$ and $\sum b_n x^n$ have convergence radius $R_a$ and $R_b$ respectively. Now we choose $0<\varepsilon<\min{\{R_a,R_b\}}$: $\sum a_n x^n$ and $\sum b_n x^n$ are uniformly convergent on $]-\varepsilon,\varepsilon[$.
Say $f_a(x) = \sum_{n=0}^{+\infty} a_nx^n = \sum_{n=0}^{+\infty} b_nx^n = f_b(x)$ $(-\varepsilon<x<\varepsilon)$. On this interval we get: $a_n = \frac{f_a^{(n)}(0)}{n!} = \frac{f_b^{(n)}(0)}{n!}$, since $0$ in always an interior point of the interval. The equality follows.
Best Answer
Assuming the convergence radii of both series are positive (as you do in your question), then one can prove the uniqueness (equality of coefficients) by differentiation. Namely, setting $x=0$ we get $a_0 = b_0$. In view of absolute convergence of the series, we can differentiate the series term by term, where the resulting series will have the same radius of convergence. Differentiating both series and setting $x=0$ implies $a_1 = b_1$. The general case of $a_n = b_n$ follows by induction on $n$.