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Let us consider the series of general term:

$$\frac{(-1)^{n-1}}{n^{1/2}}\sin(\beta \log n)$$

The question is about the convergence or the divergence of this series.

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As I mentioned to you earlier, you can find some good starting points on how to format mathematics on the site here and here. This AMS reference is very useful. If you need to format more advanced things, there are many excellent references on LaTeX on the internet, including StackExchange's own TeX.SE site. – Zev Chonoles Dec 23 '12 at 15:27
@ Zev Chonoles: I will start reading how to format mathematics. Thank you. – ZE1 Dec 23 '12 at 15:30
up vote 2 down vote accepted

We can use the complex series of the general term: $$\frac{(-1)^{n-1}}{n^{1/2}}\exp(-\beta \log n)$$ to obtain the the eta function which is analytic in the domain $Re(α+iβ)>0$. The mentioned series in the question is the imaginary part of the eta function which is convergent since the whole series is convergent. Here we have $α=0.5$.

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Use Dirichlet test for series convergence.

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It is works. Thank you very much. – ZE1 Dec 23 '12 at 16:44
@user53124: You are welcome. – Mhenni Benghorbal Dec 23 '12 at 17:01
@Mhenni Benghorbal How boundedness $\left|\sum\limits_{n=1}^{N}{{(-1)^{n-1}}\sin(\beta \log n)}\right| \leqslant M, \;\;(\forall N\in\mathbb{N})$ can be proved? – M. Strochyk Dec 23 '12 at 18:20
You know that the function $sin$ is bounded for all its real arguments. – ZE1 Dec 23 '12 at 18:30
@user53124 But sum of sines not necessary bounded – M. Strochyk Dec 23 '12 at 18:41

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