# When does $\sum_{n=0}^{\infty} \frac{2^n+n^2}{3^n+n^3}z^n$ converge?

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? – mixedmath Jun 9 '12 at 23:47
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? – mrf Jun 9 '12 at 23:51
Are you familiar with the ratio test? – Gerry Myerson Jun 10 '12 at 0:01
nour: Why are you vandalizing your own question? – Did Jun 10 '12 at 16:55
this was by mistake. I thought that I added a new question. – nour Jun 10 '12 at 21:06

## 1 Answer

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? – nour Jun 12 '12 at 12:17
we can take $a_k=(-1)^n$ and $b_k=a_n$ – nour Jun 12 '12 at 21:12
I think also that we can have convergence by Abel's test for $z=\frac{-3}{2}$ – nour Jun 13 '12 at 6:45
@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 are not satisfied with the series $a_k = (-1)^n=b_k$. – process91 Jun 13 '12 at 16:40