Does $f(z)=\displaystyle \sum_{n=0}^{\infty} \frac{2^n+n^2}{3^n+n^3}z^n$ converge for $z=\frac{-3}{2}$?
determine whether this series converges for this value of z
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calculus
real-analysis
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0To be clear, this is a subquestion of the [OP's previous queston](http://math.stackexchange.com/questions/156280/when-does-sum-n-0-infty-frac2nn23nn3zn-converge) – 2012-06-13
2 Answers
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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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0Thanks for your hints...this was helpful – 2012-06-13
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$ \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) } $