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How to show that $$ \sum_{n=2}^\infty \frac{\sin{(nx)}}{\log n} $$ not the Fourier series of any function?

I have shown that the series is convergent by Dirichlet test.

Let $a(n)=\frac{1}{\log n}$. What is $\sum (a(n))^2$, to apply Parseval's theorem?

  • 7
    See [Katznelson](http://www.amazon.com/Introduction-Harmonic-Analysis-Yitzhak-Katznelson/dp/0486633314) chapter I, 4.2. Briefly, a sine series $\sum a_n \sin{(nx)}$ is *not* a Fourier series of an integrable function if $a_n \gt 0$ and $\sum_n \frac{a_n}{n} = \infty$.2012-04-09
  • 0
    Do you mean $\sin \frac{nx}{ \log n}$ or $\frac{\sin nx}{\log n}$?2012-04-09
  • 1
    @MattN: the former has no chance of being a Fourier series...2012-04-09
  • 3
    @t.b.: google link to [An Introduction to Harmonic Analysis 4.2](http://books.google.com/books?id=gkpUE_m5vvsC&pg=PA24) (perhaps... :-))2012-04-09

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