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

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

J. M. ain't a mathematician
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JohnnyE.
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    well it is the Fourier series of $\dfrac x2$ in this interval... – Raymond Manzoni Aug 09 '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 09 '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 09 '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 09 '12 at 12:05
  • We can show the series converges uniformly on $[-\pi+\eta,\pi-\eta]$ for $0<\eta<\pi$ by Dirichlet's test. Let $x\in[-\pi+\eta,\pi-\eta]$, we have $\left|\sum_{n=1}^{m}(-1)^{n+1}\sin(nx)\right|=\left|\sum_{n=1}^{m}\sin(n(x+\pi))\right|=\left|\frac{\sin((m+1)(x+\pi))\sin(m(x+\pi))}{\sin(\frac{x+\pi}{2})}\right|\leq\frac{1}{\sin(\frac{\eta}{2})}$. – 19021605 May 01 '25 at 02:15

1 Answers1

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You can use Abel's uniform convergence test. See here. Or you can use the following theorem:

Theorem: The Fourier series of a 2π-periodic continuous and piecewise smooth function converges uniformly.

Mhenni Benghorbal
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  • To apply Abel's test we need that $g_n(x)=\sin(nx)$ is a monotone sequence of functions, i.e. $\sin(nx)\leq \sin((n+1)x)$ for all $x\in[-\pi,\pi]$ and all $n\in\mathbb{N}$, which is not the case. Hence, we can't apply Abel's test. – Philipp Oct 17 '21 at 12:01