Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
2  
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
1  
@ Zev Chonoles: I will start reading how to format mathematics. Thank you. –  ZE1 Dec 23 '12 at 15:30
add comment

2 Answers

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 $α=0.5$.

share|improve this answer
add comment

Use Dirichlet test for series convergence.

share|improve this answer
1  
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
show 4 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.