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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](http://meta.math.stackexchange.com/a/464/264) and [here](http://meta.stackexchange.com/a/70559/161783). [This AMS reference](ftp://ftp.ams.org/ams/doc/amsmath/short-math-guide.pdf) 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](http://tex.stackexchange.com/) site.2012-12-23
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    @ Zev Chonoles: I will start reading how to format mathematics. Thank you.2012-12-23

2 Answers 2

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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.2012-12-23
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    @user53124: You are welcome.2012-12-23
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    @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?2012-12-23
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    You know that the function $sin$ is bounded for all its real arguments.2012-12-23
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    @user53124 But sum of sines not necessary bounded2012-12-23
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    @M.Strochyk: Maybe you can use the proof by induction. I have no other idea.2012-12-23
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    @user53124 Right. So, in the end, do you know how to solve the question (as you accepting an answer, and your first comment, seem to indicate) or are you still lost (as your subsequent comment seems to indicate)?2012-12-25
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    @did: See my answer for this question.2012-12-25
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    @Downvoter: What's the down for?2012-12-25