Term-by-term differentiation

First we show the differentiated series has the same radius of convergence. Let $c_n = n a_n$ be the coefficients of the term-by-term derivative (shifted index). Since $n^{1/n} \to 1$ as $n \to \infty$, we have $\limsup_n |n a_n|^{1/n} = \limsup_n |a_n|^{1/n}$, and by the Cauchy–Hadamard formula the two series $\sum a_n(x-c)^n$ and $\sum n a_n (x-c)^{n-1}$ have the same radius of convergence $R$.

Next, fix any closed subinterval $[c-\rho, c+\rho]$ with $\rho < R$. On this subinterval the series of derivatives $\sum n a_n (x-c)^{n-1}$ converges uniformly, because its terms are dominated (for $n$ large) by $M \rho^{n-1}$ for constants coming from a slightly larger radius $\rho' \in (\rho, R)$, and the Weierstrass M-test applies.

Uniform convergence of the derivative series on $[c-\rho, c+\rho]$ together with pointwise convergence of the original series $\sum a_n(x-c)^n$ lets us invoke the standard theorem on differentiating a series of functions term by term: the sum $f(x)$ is differentiable on $[c-\rho,c+\rho]$ and $f'(x)$ equals the sum of the derivative series there. Since $\rho < R$ was arbitrary, this holds on all of $(c-R,\ c+R)$.