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?

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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 '14 at 3:22
Not so much haha @Braindead – user162441 Jul 10 '14 at 3:23
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 '14 at 3:31

Hint:

For which values of $y$ does

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

converge?

Now let

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

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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.

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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 '14 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 '14 at 3:36