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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2As 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
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1@ Zev Chonoles: I will start reading how to format mathematics. Thank you. – Safwane Dec 23 '12 at 15:30
2 Answers
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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 $α=0.5$.
Safwane
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Use Dirichlet test for series convergence.
Mhenni Benghorbal
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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? – M. Strochyk Dec 23 '12 at 18:20
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You know that the function $sin$ is bounded for all its real arguments. – Safwane Dec 23 '12 at 18:30
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@M.Strochyk: Maybe you can use the proof by induction. I have no other idea. – Safwane Dec 23 '12 at 19:05
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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)? – Did Dec 25 '12 at 11:13
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