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If $\lim_{t\to \infty} f(t)t=1$, i.e., $f(t)\sim t^{-1}$, then \begin{equation} {\text{ess}\sup}_{x\in [-\pi,\pi]}\sum_{t=1}^{\infty} f(t)\cos(tx)=\infty ? \end{equation} Here ${\text{ess}\sup}$ is the essential supremum. To me, it seems \begin{equation} {\text{ess}\sup}_{x\in [-\pi,\pi]}\sum_{t=1}^{\infty} t^{-1}\cos(tx)=\infty. \end{equation}

I am little confused when $\sum$ and $\lim$ come together.

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