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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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    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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    @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