[proofplan] The proof has two parts. First, we show the derived [series](/page/Series) $\sum n c_n (x-a)^{n-1}$ has the same [radius of convergence](/theorems/273) as the original by comparing their Cauchy--Hadamard formulas: the factor $n^{1/n} \to 1$ does not affect the limsup. Second, we apply the [Interchange of Limits and Derivatives](/theorems/260) on each compact sub-interval $[a - r, a + r]$ with $r < R$: the derived series converges uniformly there (by the [Radius of Convergence theorem](/theorems/262)), and the original series converges at the center $x = a$, so the hypotheses are satisfied. [/proofplan] [step:Show the derived series has radius of convergence $R$ via the Cauchy--Hadamard formula] By the Cauchy--Hadamard formula ([Theorem 262](/theorems/262)), the radius of convergence of the derived [series](/page/Series) is $1/R' = \limsup_{n \to \infty} |n c_n|^{1/n}$. We compute: \begin{align*} |n c_n|^{1/n} &= n^{1/n} \cdot |c_n|^{1/n}. \end{align*} The [sequence](/page/Sequence) $n^{1/n} \to 1$ as $n \to \infty$ (since $\log(n^{1/n}) = (\log n)/n \to 0$). For every $\varepsilon > 0$ there exists $N \in \mathbb{N}$ such that $1 \leq n^{1/n} < 1 + \varepsilon$ for all $n \geq N$. Taking the limsup and using the fact that $\limsup (a_n b_n) = \limsup(a_n) \cdot \lim(b_n)$ when $b_n$ has a finite positive [limit](/page/Limit): \begin{align*} \limsup_{n \to \infty} |n c_n|^{1/n} &= \limsup_{n \to \infty} |c_n|^{1/n} \cdot \lim_{n \to \infty} n^{1/n} = \frac{1}{R} \cdot 1 = \frac{1}{R}. \end{align*} Therefore $R' = R$. [/step] [step:Apply the interchange of limits and derivatives on each compact sub-interval] Fix any $r$ with $0 < r < R$ and work on the closed 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 $f_N'(x) = \sum_{n=1}^N n c_n (x - a)^{n-1}$. The hypotheses of the [Interchange of Limits and Derivatives](/theorems/260) require: (i) The [sequence](/page/Sequence) of derivatives $f_N'$ converges uniformly on $[a - r, a + r]$. Since the derived series has [radius of convergence](/theorems/273) $R > r$ (by the previous step), [Theorem 262](/theorems/262) (part 3) gives [uniform convergence](/page/Uniform%20Convergence) of $\sum_{n=1}^\infty n c_n (x - a)^{n-1}$ on $[a - r, a + r]$. (ii) The sequence $(f_N(c))$ converges for some $c \in [a - r, a + r]$. Taking $c = a$, we have $f_N(a) = c_0$ for all $N \geq 0$, a constant sequence. By [Theorem 260](/theorems/260), the partial sums $f_N$ converge uniformly on $[a - r, a + r]$ to a continuously differentiable [function](/page/Function), and the derivative of the limit equals the limit of the derivatives: \begin{align*} f'(x) &= \sum_{n=1}^\infty n c_n (x - a)^{n-1} \quad \text{for all } x \in [a - r, a + r]. \end{align*} [guided] We want to differentiate the [power series](/page/Power%20Series) $f(x) = \sum_{n=0}^\infty c_n (x - a)^n$ term by term. The [Interchange of Limits and Derivatives](/theorems/260) provides exactly this, but we must verify its hypotheses on each compact sub-interval $[a - r, a + r]$ with $r < R$. The first hypothesis is that the derivatives $f_N'(x) = \sum_{n=1}^N n c_n (x - a)^{n-1}$ converge uniformly on $[a - r, a + r]$. Since the derived series has the same radius of convergence $R$ (established in the previous step), and $r < R$, [Theorem 262](/theorems/262) (part 3) guarantees [uniform convergence](/page/Uniform%20Convergence) of the derived series on $[a - r, a + r]$. This is the step that requires the radius-of-convergence computation: without knowing $R' = R$, we could not assert uniform convergence of the derived series on intervals approaching the boundary. The second hypothesis is convergence of $(f_N(c))$ at some point $c \in [a - r, a + r]$. The natural choice is $c = a$: $f_N(a) = c_0 + 0 + \cdots + 0 = c_0$ for all $N \geq 0$, which converges (it is constant). Any point in the open interval of convergence would also work, but $c = a$ makes the verification elementary. With both hypotheses verified, [Theorem 260](/theorems/260) yields two conclusions: the partial sums $f_N$ converge uniformly to a continuously differentiable function on $[a - r, a + r]$, and the derivative of the limit equals the limit of the derivatives. This gives the term-by-term differentiation formula $f'(x) = \sum_{n=1}^\infty n c_n (x - a)^{n-1}$ for all $x \in [a - r, a + r]$. [/guided] [/step] [step:Extend the result to all of $(a - R, a + R)$] Since every $x \in (a - R, a + R)$ belongs to some $[a - r, a + r]$ with $|x - a| < r < R$, the formula $f'(x) = \sum_{n=1}^\infty n c_n (x - a)^{n-1}$ holds on all of $(a - R, a + R)$. [/step]