**Step 1: Periodization.** Define $\phi(t) := \sum_{n \in \mathbb{Z}} f(t + n)$. The decay condition $|f(x)| \le C(1 + |x|)^{-\alpha}$ with $\alpha > 1$ ensures absolute and [uniform convergence](/page/Uniform%20Convergence) of this [series](/page/Series) (and of the series of [derivatives](/page/Derivative)), so $\phi$ is [continuous](/page/Continuity) and periodic with period $1$. **Step 2: Fourier coefficients of $\phi$.** The $k$-th Fourier coefficient of $\phi$ (viewed as a [function](/page/Function) on $[0, 1]$) is: \begin{align*} \hat{\phi}_k = \int_0^1 \phi(t) e^{-2\pi i k t} \, d\mathcal{L}^1(t) = \int_0^1 \sum_{n \in \mathbb{Z}} f(t + n) e^{-2\pi i k t} \, d\mathcal{L}^1(t). \end{align*} By the absolute convergence, [Fubini's Theorem](/theorems/513) justifies interchanging the sum and [integral](/page/Integral): \begin{align*} \hat{\phi}_k = \sum_{n \in \mathbb{Z}} \int_0^1 f(t + n) e^{-2\pi i k t} \, d\mathcal{L}^1(t) = \sum_{n \in \mathbb{Z}} \int_n^{n+1} f(s) e^{-2\pi i k s} \, d\mathcal{L}^1(s) = \int_{\mathbb{R}} f(s) e^{-2\pi i k s} \, d\mathcal{L}^1(s) = \mathcal{F}(f)(k). \end{align*} **Step 3: Pointwise identity.** Since $\sum_{k \in \mathbb{Z}} |\mathcal{F}(f)(k)| < \infty$ by hypothesis, the [Fourier series](/page/Fourier%20Series) $g(t) := \sum_{k \in \mathbb{Z}} \hat{\phi}_k \, e^{2\pi i k t}$ converges uniformly to a continuous periodic function. Both $\phi$ and $g$ are continuous with the same Fourier coefficients, so $\phi = g$ on $[0, 1]$ (equal in $L^2([0, 1])$ and both continuous). Evaluating at $t = 0$: \begin{align*} \sum_{n \in \mathbb{Z}} f(n) = \phi(0) = g(0) = \sum_{k \in \mathbb{Z}} \mathcal{F}(f)(k). \end{align*}