How may one deal with a divergent Fourier series for example

$$ f(x) := \sum_{n=0}^{\infty} \frac{\cos(n\theta+\alpha)}{ n^{a}}$$ or

$$ g(x) := \sum_{n=0}^{\infty} \frac{\sin(\log (n\theta)+\beta)}{ n^{a}} \ ? $$

Do we trucante each divergent series upt to a certain number $N = N(x)$ but how to get this number $N$? Or perhaps apply Cesaro summation formula to the means $ m_{K}= \sum_{i=0}^{K}\frac{sin(i+\alpha)}{i^{\alpha}}$? However in this latter case, how do we apply the formula for the sum $ S(x) = \frac{ m_1(x)+ m_2(x)+ \cdots +m_N}{N} $ in the limit $ N \rightarrow \infty $


Your first series (start with $n=1$) can be written in the form $$f(x)=\sum_{n=1}^\infty {e^{i(n\theta+\alpha)}+e^{-i(n\theta+\alpha)}\over 2n^\alpha}={e^{i\alpha}\over 2}\sum_{n=1}^\infty {e^{i n\theta}\over n^\alpha}+{e^{-i\alpha}\over 2}\sum_{n=1}^\infty {e^{-i n\theta}\over n^\alpha}\ .$$ The two series on the right hand are absolutely convergent when $\alpha>1$, and are convergent when $0<\alpha\leq1$ and $\theta\ne 0\ (2\pi)$, by Dirichlet's test. When $\alpha\leq0$ they are divergent; you can't do anything about it.

Your second series is not a Fourier with respect to the variable $\theta$ and cannot be converted into a Fourier series of some auxiliary variable $\tau$ by means of the substitution $\theta:=e^\tau$, or the like. All we can do is judge whether the series as written is convergent. When $\alpha>1$ it is certainly absolutely convergent, and it may be convergent also for some $\alpha>0$; but this would be difficult to prove.

  • $\begingroup$ Isn't $\sum e^{in\theta}n^{-\alpha}$ convergent for all $\alpha>0$, by the Dirichlet test? $\endgroup$ – user31373 Sep 26 '12 at 13:54
  • $\begingroup$ @LVK: You are of course right. I've corrected that. $\endgroup$ – Christian Blatter Sep 26 '12 at 14:36

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