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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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You're given

$$f(z)=\displaystyle \sum_{n=0}^{\infty} \frac{2^n+n^2}{3^n+n^3}z^n$$

A sensible solution would be using Cauchy's Root test. We want to find

$$\lim\limits_{n\to\infty}\left(\frac{2^n+n^2}{3^n+n^3}\right)^{1/n} =$$

$$=\lim\limits_{n\to\infty}\frac 2 3\left(\frac{1+n^2/2^n}{1+n^3/3^n}\right)^{1/n} =$$

$$=\frac 2 3\left(\frac{1+0}{1+0}\right)^{0}=\frac 2 3 $$

Then the sum converges for $|z|<\dfrac 3 2 $

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    I think also that for z=$\frac{-3}{2}$ the series converges by Abel's test. right?2012-06-12
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    we can take $a_k=(-1)^n$ and $b_k=a_n$2012-06-12
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    I think also that we can have convergence by Abel's test for $z=\frac{-3}{2}$2012-06-13
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    @nour I see you have asked another question which addresses $z=\tfrac {-3} 2$, but I just wanted to point out that the criteria for [Abel's test](http://en.wikipedia.org/wiki/Abel's_test) are not satisfied with the series $a_k = (-1)^n=b_k$.2012-06-13