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Where is the series $\displaystyle\sum_{n=1}^{\infty}\dfrac{\cos(nx)}{n}$ pointwise convergent? I tried to apply the Dirichlet's test but I couldn't.

hmakholm left over Monica
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Chilote
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2 Answers2

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Where did you find trouble in applying it? Hint Find a closed form for $$\sum_{k=0}^n\cos kx$$ that let's you decided when this sum is bounded. A good idea might be to note this is $$\Re\left(\sum_{k=0}^n e^{ikx}\right)$$

Pedro
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  • If $x=0$ the partial sum $\sum_{k=0}^{\infty}e^{ikx}$ is divergent but I don't have any clue when $x\neq 0$. – Chilote Apr 21 '14 at 03:14
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    @Chilote It's a geometric series with ratio $q=e^{ix}$. – Pedro Apr 21 '14 at 17:36
  • But in that case the ratio is of modulus 1 and a geometric series only converge when the ratio is strictly smaller than 1. – Chilote Apr 21 '14 at 20:30
  • @Chilote We're not looking for convergence, but boundedness. – Pedro Apr 21 '14 at 22:19
  • Oh! that's true! So this settles the convergence of the series for all $x\neq 2k\pi$. And what about the values of the series? Is there any familiar function which represent the values of the series? – Chilote Apr 22 '14 at 01:37
  • @Chilote Yes, it's the Fourier series of... (Claude wasn't entirely incorrect) – Pedro Apr 22 '14 at 01:47
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Hint

Following Pedro Tamaroff, may be, you could consider two summations $$I=\displaystyle\sum_{n=1}^{\infty}\dfrac{\cos(nx)}{n}$$ $$J=\displaystyle\sum_{n=1}^{\infty}\dfrac{\sin (nx)}{n}$$ Then $$I+iJ=\displaystyle\sum_{n=1}^{\infty}\dfrac{e^{inx}}{n}=\displaystyle\sum_{n=1}^{\infty}\dfrac{y^{n}}{n}=-\log (1-y)$$ where $y=e^{ix}$ and use the real part of the complex logarithm.

Claude Leibovici
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