> [!NOTE] Statement 1 (Pointwise Convergent Series of Continuously Differentiable Functions Permits Term-By-Term Differentiation When Series of Derivatives Converges Uniformly) > Let $(f_{k}),f_{k}:[a,b]\to \mathbb{R}$ be a sequence of [[Continuous Differentiability|continuously differentiable]] functions such that $\sum_{k=1}^{\infty} f_{k}$ [[Pointwise Convergence of Series of Real Functions|converges pointwise]]. Assume that $\sum_{k=1}^{n} f_{k}'$ [[Uniform Convergence of Series of Real Functions|converges uniformly]]. Then $\left( \sum_{k=1}^{n} f_{k}(x) \right)' = \sum_{k=1}^{\infty} f_{k}'(x)$that is, the series is [[Fréchet Differentiation|differentiable]] and can be differentiated term-by-term. ###### Proof The statement of [[Uniform Convergence and Differentiability]] is that if $S_{n}$ is $C^1,$ $S_{n}\to S,$ $S_{n}' \rightrightarrows g$ then $S\in C^1$ and $S'=g$. Let $S_{n}:=\sum_{k=1}^{n} f_{k}.$ Then indeed $S_{n}\in C^1$ since each we assume that each $f_{k}$ is $C^1$ and [[Derivative of Sum of Differentiable Real Functions|derivative is additive]]. By our assumption, we have also $S_{n}$ that converges pointwise to $S=\sum_{k=1}^\infty f_{k}$ and $S_{n}'$ converges uniformly, to say $g.$ Therefore, as above, $S\in C^{1}$ and $S_{n}' \rightrightarrows S'.$ This means $\lim_{ n \to \infty } S_{n}' = S' = \left( \sum_{k=1}^\infty f_{k}(x) \right)' $but since $S_{n}'=\left( \sum_{k=1}^n f_{k} \right)'=\sum_{k=1}^n f_{k}'$ we get $\sum_{k=1}^\infty f_{k}'(x)= \left( \sum_{k=1}^\infty f_{k}(x) \right)'.$ $\blacksquare$