[proofplan] Part (1) follows from the root test: $\limsup |nc_n|^{1/n} = \limsup |c_n|^{1/n}$ since $n^{1/n} \to 1$, so the derived [series](/page/Series) has the same [radius of convergence](/theorems/273). Part (2) applies the [Interchange of Limits and Derivatives](/theorems/260) on closed sub-intervals $[a-r, a+r]$ with $r < R$, using [uniform convergence](/page/Uniform%20Convergence) of the derived series from Part (1) and convergence of the original series at $x = a$. [/proofplan] [step:Show the derived series has radius of convergence $R$] Let $\alpha = \limsup_{n \to \infty} |c_n|^{1/n}$, so $R = 1/\alpha$. The $n$-th root of the absolute value of the $n$-th coefficient of the derived [series](/page/Series) is $|nc_n|^{1/n} = n^{1/n} |c_n|^{1/n}$. Since $\lim_{n \to \infty} n^{1/n} = 1$ (as $\log(n^{1/n}) = (\log n)/n \to 0$), the product rule for limsup with a convergent factor gives \begin{align*} \limsup_{n \to \infty} |nc_n|^{1/n} &= 1 \cdot \alpha = \alpha. \end{align*} Therefore the derived series has [radius of convergence](/theorems/273) $1/\alpha = R$. [/step] [step:Verify the hypotheses of the interchange theorem on each compact sub-interval] Fix any $r$ with $0 < r < R$. On the interval $[a-r, a+r]$, define $f_N(x) = \sum_{n=0}^N c_n(x-a)^n$. Each $f_N$ is a polynomial, hence [continuously](/page/Continuity) [differentiable](/page/Derivative), with $f_N'(x) = \sum_{n=1}^N nc_n(x-a)^{n-1}$. By the previous step, the derived [series](/page/Series) has radius of convergence $R > r$. By Part (3) of the [Radius of Convergence theorem](/theorems/265), the derived series converges uniformly on $[a-r, a+r]$, i.e., $f_N' \to g$ uniformly where $g(x) = \sum_{n=1}^\infty nc_n(x-a)^{n-1}$. The original series converges at $x = a$ (giving $f(a) = c_0$), so convergence at a single point is satisfied. By the [Interchange of Limits and Derivatives](/theorems/260), $f$ is differentiable on $(a-r, a+r)$ with $f' = g$. [/step] [step:Extend to all of $(a-R, a+R)$] Since $r < R$ was arbitrary and every $x \in (a-R, a+R)$ lies in some $[a-r, a+r]$ with $|x - a| < r < R$, the formula \begin{align*} f'(x) &= \sum_{n=1}^\infty nc_n(x-a)^{n-1} \end{align*} holds on all of $(a-R, a+R)$. [/step]