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.

How to find the values for which $z$ converges, $z\in\mathbb{C}$, in the serie $$\sum_{n=1}^{\infty}\frac{1}{(1+|z|^{2})^{n}}$$ I know I have to use the convergence radius expression, but what I suppose to do here? I mean, as $$R=\lim_{n\to\infty}\left|\frac{a_{n}}{a_{n+1}}\right|$$ what is $a_{n}$ in this case?

share|improve this question
    
The formula you are writing is for $\sum_{n=1}^n a_n z^n$. You maybe able to rewrite your series into this form, but there is an easier way: Are you familiar with the ratio test? –  Braindead Jul 10 at 3:22
    
Not so much haha @Braindead –  user162441 Jul 10 at 3:23
1  
The following link has some nice examples. sosmath.com/calculus/radcon/radcon02/radcon02.html –  Braindead Jul 10 at 3:25
    
Sorry, in my first comment, it should say $\sum_{n=1}^\infty a_n z^n$ (or more appropriately for this case: $\sum_{n=1}^\infty a_n |z|^n$, since your expression depends on $|z|$ and not so much on $z$.) –  Braindead Jul 10 at 3:31

2 Answers 2

up vote 5 down vote accepted

Hint:

For which values of $y$ does

$$ \sum_{n=1}^{\infty} y^n $$

converge?

Now let

$$ y = \frac{1}{1+|z|^2}. $$

share|improve this answer

Hint This is not really a question about complex numbers. Think geometric series.

For the question as it stands now, you do not have to use the convergence radius expression. But to answer your question about $a_n$, if you will use the Ratio Test then you should use $a_n=\frac{1}{(1+|z|^2)^n}$. Note that our series is not a power series in the usual sense. So formulas you may remember about radius of convergence need not apply.

share|improve this answer
    
I know what you mean, but I think your hint is a bit confusing for the questioner, since your "formula" for $a_n$ does not fit into the formula for Radius of convergence that user162441 put up. –  Braindead Jul 10 at 3:29
    
Thank you, I have reworded. The warning I had given about it not being a power series was probably too gnomic. –  André Nicolas Jul 10 at 3:36

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.