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Does anyone know of a convergence test for a complex series of the form

$$\sum_n a_n \cdot \exp(i \cdot b_n)$$

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The particular series I need to understand has a_n going to zero as n goes to infinity, but it fails the absolute convergence test. However numerically I do find it to converge. It should have something to do with convergence of Fourier series (since $\ \exp(i\cdot t)\ =\ \cos(t)\,+\,i\cdot \sin(t)$).

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    $\begingroup$ This seems very vague to me. Why don't you describe the particular series? $\endgroup$
    – Yemon Choi
    Commented Sep 19, 2014 at 21:55
  • $\begingroup$ I need something stronger than the Dirichlet test, since my series fails it, but still converges! Are such examples known? $\endgroup$
    – André LeClair
    Commented Sep 20, 2014 at 17:35
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    $\begingroup$ Somehow, this got put on hold, but whoever suggested summation by parts, thank you! It did the trick. $\endgroup$
    – André LeClair
    Commented Oct 15, 2014 at 18:14

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If your $a_n$s are nonnegative, decrease monotonically, and approach zero, and the $b_n$s are such that $\sum e^{i x b_n}$ remains bounded, then Dirichlet's test would apply here and give you convergence. If the $b_n$'s are an arithmetic sequence, as in the case of Fourier series, then you have the second of these criteria (for $x\neq0$).

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  • $\begingroup$ Yoav, thanks. But my series fails the Dirichlet test but still converges! Any ideas? $\endgroup$
    – André LeClair
    Commented Sep 20, 2014 at 17:41
  • $\begingroup$ I refined the question in my next post. $\endgroup$
    – André LeClair
    Commented Sep 20, 2014 at 21:09
  • $\begingroup$ Apparently the Dirichlet test is not an "if and only" statement. My series fails it, but still converges. Is that right? $\endgroup$
    – André LeClair
    Commented Sep 20, 2014 at 21:21

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