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For what values of $z \in \mathbb{C}$ does the following series converge:

$$\displaystyle \sum_{n=0}^{\infty} \frac{2^n+n^2}{3^n+n^3}z^n\quad ?$$

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    I bet there are some missing parentheses?2012-06-09
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    Do you mean $$\sum_{n=0}^\infty \frac{2^n+n^2}{3^n+n^3}\,z^n\:?$$And as with your other question, what have you tried?2012-06-09
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    Are you familiar with the ratio test?2012-06-10
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    nour: Why are you vandalizing your own question?2012-06-10
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    this was by mistake. I thought that I added a new question.2012-06-10
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    @nour: I have put your original question back. Please ask a new question [by going here](http://math.stackexchange.com/questions/ask) (or go to the "Ask Question" button on the upper right of the site).2012-06-11
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    Try using the root test, combined with the fact that $n^k/a^n \to 0$ for $k \ge 0$ and $a > 1$ as n gets large.2012-06-11
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    Hint: note that $\dfrac{2^n+n^2}{3^n+n^3} - \dfrac{2^n}{3^n} = O\left(\dfrac{n^3}{3^n}\right)$. Your series converges whenever $\sum_{n=0}^\infty \left(\dfrac{2z}{3}\right)^n$ converges.2012-06-11

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