# Convergence of: $\sum\limits_{n=1}^{\infty} \frac{(-1)^{n+1}\sin(nx)}{n}$

Need help with checking: $\sum\limits_{n=1}^{\infty} \frac{(-1)^{n+1}\sin(nx)}{n}$

for point-wise convergence and uniform convergence of: ${-\pi} \leq x \leq {\pi}$.

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well it is the Fourier series of $\dfrac x2$ in this interval... –  Raymond Manzoni Aug 9 '12 at 11:48
yes that is true, but as i understand it the series doesn't converge to $\frac{x}{2}$ at every point in the interval –  JohnnyE. Aug 9 '12 at 11:52
in $(-\pi,\pi)$ it will be $\frac x2$ (repeated every $2\pi$). At multiples of $\pi$ it will be $0$ ($0=\frac {\pi/2-\pi/2}2$). Did you get something else? –  Raymond Manzoni Aug 9 '12 at 11:56
(to detail a little) the values at $-\pi$ and $\pi$ will be $0$ because of the jump from $-\frac {\pi}2$ to $+\frac {\pi}2$. At a jump discontinuity of first order the value is at the middle ($\frac {-\pi/2+\pi/2}2$). (my first comment should have been $\frac x2$ in $(-\pi,\pi)$ and not $[-\pi,\pi]$...) –  Raymond Manzoni Aug 9 '12 at 12:05