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I'd like to prove the error estimate for the factional part of $x$ as given above.

To begin with, we apply the Abel's formula and get

$$(1):\left|\sum_{n=N+1}^M \frac{\sin(2\pi n x) }{n}\right|=\left|\frac {\sum_{n=N+1}^{M} \sin(2\pi n x)}{M}+ \\ \sum_{k=N+1}^{M-1} \left (\sum_{k=1}^n \sin(2\pi k x) \right)(\frac{1}n- \frac{1}{n+1}) \right|$$

Moreover, we have $$(2): \left|\sum_{n=N+1}^M \sin(2\pi n x) \right|=\left| \frac{\cos(2\pi Nx)-\cos(2\pi Mx)}{2\sin(\pi x)} \right| \\= \left | \frac{\sin(\pi (N-M)x)\sin(\pi (M+N)x)}{\sin(\pi x)} \right |.$$

I hope my calculation is correct. If so, how do I conclude from those equations that (as $M \to \infty$) the error is $O((1+\|x\|N)^{-1})$, as $x$ approaches any integer and $N\to \infty$?

I think I am able to see that when $x$ is a fix number away from any integer, then the error is $O(1/N)$. However, I don't know how to handle this when $x$ is considered.

There is solution to the same question with very different method https://math.stackexchange.com/a/57207/185631

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  • When the coefficients don't have a such a simple closed-form use the Dirichlet kernel $f(x)-\sum_{n=-N}^N (\int_0^1 f(y)e^{-2i \pi ny}dy) e^{2i \pi nx}= f(x)-\int_{-1/2}^{1/2} f(x-y) \frac{\sin\left(\left(N +1/2\right) 2\pi y \right)}{2\pi\sin(2\pi y/2)} dy$. When $f$ is $C^1$ around $x$ we can use $\frac{\sin\left(\left(N +1/2\right) 2\pi y \right)}{2\pi\sin(2\pi y/2)} \approx \frac{\sin\left(\left(N +1/2\right) 2\pi y \right)}{2\pi 2\pi y/2}$ and integrate by parts to use that the primitive of $\frac{\sin\left(\left(N +1/2\right) 2\pi y \right)}{2\pi 2\pi y/2}$ converges to $sign(y)$ – reuns Aug 23 '19 at 08:14

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