A simple but useful estimate

Theorem. Let $f : \mathbb{R} \to \mathbb{R}$ be a continuous, integrable function of bounded variation. Then $\sum_{n\in\mathbb{Z}} f(n)$ converges absolutely, and we have $$ \left| \sum_{n\in \mathbb{Z}} f(n) - \int_{\mathbb{R}} f(x) \, \mathrm{d}x \right| \leq \frac{1}{2} V(f), $$ where $V(f) = \int_{\mathbb{R}} |\mathrm{d}f(x)|$ denotes the total variation of $f$ over $\mathbb{R}$.

Proof. Let $\tilde{B}(x) = x - \lfloor x \rfloor - \frac{1}{2}$. Then performing integration by parts, for $a \le b$, \begin{align*} \int_{a}^{b} f(x) \, \mathrm{d}x - \sum_{n \in [a, b]\cap\mathbb{Z}} f(n) &= \int_{[a, b]} f(x) \, \mathrm{d}\tilde{B}(x) \\ &= [f(x)\tilde{B}(x)]_{a^-}^{b} - \int_{[a,b]} \tilde{B}(x) \, \mathrm{d}f(x). \end{align*} Using the fact that $|\tilde{B}(x)| \leq \frac{1}{2}$, it therefore follows that \begin{align*} \left| \int_{a}^{b} f(x) \, \mathrm{d}x - \sum_{n \in [a, b]\cap\mathbb{Z}} f(n) \right| \leq \frac{|f(a)| + |f(b)|}{2} + \frac{1}{2} \int_{[a,b]} \, |\mathrm{d}f(x)| \end{align*} From the assumption, it is easy to check that $f(x) \to 0$ as $|x| \to \infty$. So, the above estimate shows that $\sum_{n\in\mathbb{Z}} f(n)$ satisfies the Cauchy criterion and hence converges as $a \to -\infty$ and $b \to \infty$. Moreover, noting that $|f|$ also satisfies the assumption in plcae of $f$, we find that this sum converges absolutely. Finally, passing to the limit as $[a,b] \uparrow \mathbb{R}$ proves the desired bound. $\square$