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Does $$f(z)=\displaystyle \sum_{n=0}^{\infty} \frac{2^n+n^2}{3^n+n^3}z^n$$ converge for $z=\frac{-3}{2}$?

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In general, text should not go between dollars. –  Julián Aguirre Jun 13 '12 at 15:37
    
Is this a homework question? –  Théophile Jun 13 '12 at 15:40
    
no, I'm studying for GRE test.... I'm trying using Abel's test but I need to clarify this answer –  nour Jun 13 '12 at 15:41
    
it would be better if you put your solution here and ask for the clarification –  Ilya Jun 13 '12 at 15:42
    
To be clear, this is a subquestion of the OP's previous queston –  mixedmath Jun 13 '12 at 15:47

2 Answers 2

Hint (assuming this is homework):

Consider the terms $$\frac{2^n+n^2}{3^n+n^3}\left(\frac{-3}{2}\right)^n.$$

Can you find the limit of this expression as $n\to\infty$?

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I Know that this series converges for $|z|<\frac{3}{2}$ by ratio test...and diverges for $z=\frac{3}{2}$ using limit of $\frac{a_n}{a_{n+1}}$.but what about $z=\frac{-3}{2}$ –  nour Jun 13 '12 at 15:45
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Then ask: Can the series converge if the terms do not converge to zero? –  Old John Jun 13 '12 at 15:55
    
so series diverges for $z=\frac{-3}{2}$ –  nour Jun 13 '12 at 15:58
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It certainly cannot converge. –  Old John Jun 13 '12 at 16:00
    
Thanks for your hints...this was helpful –  nour Jun 13 '12 at 16:01

$$ \frac{2^n + n^2}{3^n + n^3} z^n = \frac{2^n}{3^n + n^3} z^n + \frac{n^2}{3^n + n^3} z^n $$ $$ \frac{n^2}{3^n + n^3} z^n < n^2 \left ( \frac{z}{3} \right )^n = \frac{n^2 }{(-2)^n } \text{ Which converges from Ratio test }$$ $$ \frac{2^n}{3^n + n^3} z^n = \left ( \frac{2}{3} z\right )^n \frac{1}{1 + \frac{n^3}{3^n}} \text{ which is } (-1)^n \text{ for } n \rightarrow \infty \text{ (It does not converge) } $$

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