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I have calculated the Fourier Series of $g\left(x\right)=x$ on $\left(-\pi,\pi\right]$ extended periodically to $\mathbb{R}$ to be $$g\left(x\right)=2\sum^{\infty}_{n=1}\dfrac{\left(-1\right)^{n+1}}{n}\sin\left(nx\right)$$

I now need to determine and explain whether the Fourier series converges at $x=\pi$ but I'm not sure how.

If it involves finding the left/right limits/derivatives, I'm not sure how to find these for a Fourier Series.

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    $\begingroup$ What is $\sin(n\pi)$? $\endgroup$ – David C. Ullrich May 24 '16 at 18:57
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for $x=\pi$, the series is $$ \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} \sin(n\pi) = \sum_{n=1}^{\infty} 0 = 0 $$

so the series converges to 0 at $\pi$. We expect this because 0 is the midpoint between $g(-\pi)$ and $g(\pi)$

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