The smoothed sum $\sum a(n)\widetilde{1_\varepsilon}(n/N)$ is within $O(\varepsilon N\log N)$ of $\sum_{n<N}a(n)$

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The smoothed sum $\sum a(n)\widetilde{1_\varepsilon}(n/N)$ is within $O(\varepsilon N\log N)$ of $\sum_{n<N}a(n)$ · Prove2Me