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Why the series is divergent, but the equation holds? $$\sum\limits_{i=1}^{\infty }{\sin kx}=\frac{1}{2}\cot \left( \frac{x}{2} \right)$$

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This question is incorrect. What kind of equality is this? Approximately everywhere, in distributions, pointwise? –  userNaN Feb 2 '13 at 10:41
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The LHS is defined on $2\pi\mathbb{Z}$ while the RHS isn't! –  Mercy Feb 2 '13 at 12:37

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Because you assumed that the series from which you derived it could be differentiated, when it could not.

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Please explain a bit more... –  vonbrand Feb 2 '13 at 13:53
    
@vonbrand: The Fourier series is not necessarily convergent at individual points; rather, it converges with respect to the $L^2(-\pi,\pi)$ metric. So even though a Fourier series representing a function does converge, the FS representing the derivative of that function is not guaranteed to converge. So there are additional conditions on the function, i.e., the piecewise continuity of the original function. See, for example, tutorial.math.lamar.edu/Classes/DE/… –  Ron Gordon Feb 2 '13 at 17:15
    
that I (mostly) knew, but I'm at a loss at the series you claim OP differentiated. –  vonbrand Feb 2 '13 at 17:21
    

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