[proofplan] The hypothesis $\sum |n|^k|\hat{f}(n)| < \infty$ ensures that the formally differentiated series $\sum (in)^j\hat{f}(n)\,e^{inx}$ converges absolutely and uniformly for $j \leq k$ by the Weierstrass M-test. Uniform convergence of the original series identifies $f$ with a continuous function, and the theorem on term-by-term differentiation of uniformly convergent series of smooth functions gives $f^{(j)}(x) = \sum (in)^j\hat{f}(n)\,e^{inx}$. [/proofplan] [step:Establish absolute and uniform convergence of the series $\sum (in)^j\hat{f}(n)\,e^{inx}$ for $j \leq k$] For $0 \leq j \leq k$, the terms satisfy: \begin{align*} |(in)^j\hat{f}(n)\,e^{inx}| = |n|^j|\hat{f}(n)| \leq |n|^k|\hat{f}(n)| \end{align*} for $|n| \geq 1$ and $j \leq k$. Since $\sum_{n \in \mathbb{Z}} |n|^k|\hat{f}(n)| < \infty$ by hypothesis, the Weierstrass M-test gives absolute and uniform convergence of $g_j(x) := \sum_{n \in \mathbb{Z}} (in)^j\hat{f}(n)\,e^{inx}$ for each $j \leq k$. In particular, each $g_j$ is continuous. [/step] [step:Identify $f$ with $g_0$ and perform term-by-term differentiation] For $j = 0$: the series $g_0(x) = \sum\hat{f}(n)\,e^{inx}$ converges uniformly. By [Completeness](/theorems/585), the Fourier series of $f \in L^2(\mathbb{T})$ converges to $f$ in $L^2$. Since the series also converges uniformly, the uniform limit equals $f$ almost everywhere. After modification on a null set, $f = g_0$ is continuous. For $j = 1$: the series $g_0$ converges uniformly and the formally differentiated series $g_1(x) = \sum in\hat{f}(n)\,e^{inx}$ also converges uniformly. By the theorem on term-by-term differentiation of uniformly convergent series of $C^1$ functions (each $e^{inx}$ is $C^\infty$), $g_0' = g_1$. Iterating: $g_0^{(j)} = g_j$ for $0 \leq j \leq k$. [/step] [step:Conclude $f \in C^k(\mathbb{T})$ with uniformly convergent Fourier series for all derivatives] The function $f$ agrees (after modification on a null set) with $g_0 \in C^k(\mathbb{T})$, and: \begin{align*} f^{(j)}(x) = g_j(x) = \sum_{n \in \mathbb{Z}} (in)^j\hat{f}(n)\,e^{inx} \quad \text{uniformly, for } 0 \leq j \leq k. \end{align*} [/step]