[proofplan]
We first show that the derived series has the same [radius of convergence](/theorems/265) $R$ via the [Cauchy-Hadamard Formula](/theorems/203), using the fact that $n^{1/n} \to 1$. We then prove differentiability at each point $x_0 \in (c - R, c + R)$ by bounding the difference quotient error $|[f(x) - f(x_0)]/(x - x_0) - g(x_0)|$ with a convergent series multiplied by $|x - x_0|$, using two applications of the mean value theorem.
[/proofplan]
[step:Show the derived series has the same radius of convergence]
Let $R'$ denote the [radius of convergence](/theorems/262) of $\sum_{n=1}^{\infty} n a_n (x-c)^{n-1}$. The coefficients of the derived series, viewed as a [power series](/page/Power%20Series) in $(x - c)$, are $b_n = (n+1)a_{n+1}$ for $n \geq 0$. Equivalently, we compute the $\limsup$ of the $n$-th roots of $|n a_n|$:
\begin{align*}
\limsup_{n \to \infty} |n a_n|^{1/n} = \limsup_{n \to \infty} n^{1/n} \cdot |a_n|^{1/n}.
\end{align*}
Since $n^{1/n} \to 1$ as $n \to \infty$, the factor $n^{1/n}$ does not affect the $\limsup$:
\begin{align*}
\limsup_{n \to \infty} |n a_n|^{1/n} = 1 \cdot \limsup_{n \to \infty} |a_n|^{1/n}.
\end{align*}
The [Cauchy-Hadamard Formula](/theorems/203) gives $R' = 1 / \limsup_{n \to \infty} |n a_n|^{1/n} = 1 / \limsup_{n \to \infty} |a_n|^{1/n} = R$.
[guided]
Why does multiplying coefficients by $n$ not change the [radius of convergence](/theorems/273)? The radius is determined by the $\limsup$ of the $n$-th roots of the absolute values of the coefficients. The $n$-th root $|n a_n|^{1/n} = n^{1/n} |a_n|^{1/n}$, and the factor $n^{1/n}$ converges to $1$. More precisely, $\log(n^{1/n}) = (\log n)/n \to 0$, so $n^{1/n} = e^{(\log n)/n} \to e^0 = 1$. A sequence converging to $1$ does not affect the $\limsup$ of the product:
\begin{align*}
\limsup_{n \to \infty} n^{1/n} \cdot |a_n|^{1/n} = \limsup_{n \to \infty} |a_n|^{1/n},
\end{align*}
because for any $\varepsilon > 0$, we have $(1 - \varepsilon)|a_n|^{1/n} \leq n^{1/n}|a_n|^{1/n} \leq (1 + \varepsilon)|a_n|^{1/n}$ for all sufficiently large $n$. The [Cauchy-Hadamard Formula](/theorems/203) then gives $R' = R$. The same argument applies to higher derivatives: the series $\sum n(n-1) a_n (x - c)^{n-2}$ also has radius of convergence $R$, because $(n(n-1))^{1/n} \to 1$ as well.
[/guided]
[/step]
[step:Bound the difference quotient error using the mean value theorem]
Fix $x_0 \in (c - R, c + R)$ and choose $r$ with $|x_0 - c| < r < R$. Define the function
\begin{align*}
g: (c - R, c + R) &\to \mathbb{R} \\
x &\mapsto \sum_{n=1}^{\infty} n a_n (x - c)^{n-1}.
\end{align*}
For $x \in (c - r, c + r)$ with $x \neq x_0$, the difference quotient error is
\begin{align*}
\frac{f(x) - f(x_0)}{x - x_0} - g(x_0) = \sum_{n=1}^{\infty} a_n \left(\frac{(x-c)^n - (x_0-c)^n}{x - x_0} - n(x_0 - c)^{n-1}\right).
\end{align*}
By the [Mean Value Theorem](/theorems/186) applied to the function $t \mapsto t^n$ on the interval between $x - c$ and $x_0 - c$, for each $n \geq 1$ there exists $\xi_n$ between $x - c$ and $x_0 - c$ such that
\begin{align*}
\frac{(x-c)^n - (x_0-c)^n}{x - x_0} = n \xi_n^{n-1}.
\end{align*}
Applying the Mean Value Theorem again, now to $t \mapsto n t^{n-1}$, the difference $|n \xi_n^{n-1} - n(x_0 - c)^{n-1}|$ is bounded by $n(n-1) r^{n-2} |\xi_n - (x_0 - c)|$, and since $|\xi_n - (x_0 - c)| \leq |x - x_0|$:
\begin{align*}
\left| \frac{(x-c)^n - (x_0-c)^n}{x - x_0} - n(x_0 - c)^{n-1} \right| \leq n(n-1) r^{n-2} |x - x_0|.
\end{align*}
Summing over $n$:
\begin{align*}
\left|\frac{f(x) - f(x_0)}{x - x_0} - g(x_0)\right| \leq |x - x_0| \sum_{n=2}^{\infty} n(n-1)|a_n| r^{n-2}.
\end{align*}
[guided]
The core idea is to control how fast the difference quotient of each monomial $(x - c)^n$ approaches its derivative $n(x_0 - c)^{n-1}$, using two applications of the [Mean Value Theorem](/theorems/186).
**First MVT application.** For the function $\phi_n(t) = t^n$, the Mean Value Theorem gives
\begin{align*}
\frac{(x-c)^n - (x_0-c)^n}{x - x_0} = \frac{\phi_n(x-c) - \phi_n(x_0-c)}{(x-c) - (x_0-c)} = n\xi_n^{n-1}
\end{align*}
for some $\xi_n$ between $x - c$ and $x_0 - c$. Since both $|x - c|$ and $|x_0 - c|$ are at most $r$, we have $|\xi_n| \leq r$.
**Second MVT application.** For the function $\psi_n(t) = nt^{n-1}$, we bound the difference $|\psi_n(\xi_n) - \psi_n(x_0 - c)|$. Since $\psi_n'(t) = n(n-1)t^{n-2}$, the Mean Value Theorem gives
\begin{align*}
|n\xi_n^{n-1} - n(x_0 - c)^{n-1}| \leq \sup_{|t| \leq r} |n(n-1)t^{n-2}| \cdot |\xi_n - (x_0 - c)| = n(n-1)r^{n-2} |\xi_n - (x_0 - c)|.
\end{align*}
Since $\xi_n$ lies between $x - c$ and $x_0 - c$, we have $|\xi_n - (x_0 - c)| \leq |(x - c) - (x_0 - c)| = |x - x_0|$. Combining:
\begin{align*}
\left|\frac{(x-c)^n - (x_0-c)^n}{x - x_0} - n(x_0 - c)^{n-1}\right| \leq n(n-1)r^{n-2}|x - x_0|.
\end{align*}
Multiplying by $|a_n|$ and summing over $n \geq 2$ (the $n = 1$ term contributes zero since $1 \cdot 0 \cdot r^{-1} = 0$):
\begin{align*}
\left|\frac{f(x) - f(x_0)}{x - x_0} - g(x_0)\right| \leq |x - x_0| \sum_{n=2}^{\infty} n(n-1)|a_n| r^{n-2}.
\end{align*}
[/guided]
[/step]
[step:Verify the bounding series converges and conclude differentiability]
The series $\sum_{n=2}^{\infty} n(n-1)|a_n| r^{n-2}$ has radius of convergence $R$ by the same $n$-th root argument applied twice: $\limsup_{n \to \infty} (n(n-1)|a_n|)^{1/n} = \limsup_{n \to \infty} |a_n|^{1/n}$ since $(n(n-1))^{1/n} \to 1$. Since $r < R$, the series converges. Let
\begin{align*}
C := \sum_{n=2}^{\infty} n(n-1)|a_n| r^{n-2} < \infty.
\end{align*}
Given $\epsilon > 0$, set $\delta = \epsilon / C$ (if $C = 0$ any $\delta$ works). For $0 < |x - x_0| < \delta$ with $x \in (c - r, c + r)$:
\begin{align*}
\left|\frac{f(x) - f(x_0)}{x - x_0} - g(x_0)\right| \leq |x - x_0| \cdot C < \delta \cdot C = \epsilon.
\end{align*}
Therefore $f'(x_0) = g(x_0) = \sum_{n=1}^{\infty} n a_n (x_0 - c)^{n-1}$. Since $x_0 \in (c - R, c + R)$ was arbitrary, the identity $f'(x) = \sum_{n=1}^{\infty} n a_n (x - c)^{n-1}$ holds on the entire interval of convergence.
[/step]
