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I'm trying to find the radius of convergence for $\sum \limits_{n=0}^\infty(-1)^n9^nx^{2n}$. And I'm honestly new to this topic. So it would be cool if someone could tell me if my steps do make sense.

By using the root test I get:

$$\lim_{n \to \infty} \sqrt[n] {|(-1)^n9^nx^{2n}|} $$ $$=\lim_{n \to \infty} \sqrt[n] {|(-1)|^n} \cdot 9x^2$$ $$=\lim_{n \to \infty} 9x^2 = \infty$$

$$\Rightarrow R= 0$$

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    I can honestly tell you that it looks fine. – 2017-01-11
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    @fatalError Check your computations. Note that $\lim_{n\to\infty}9x^2=9x^2$ and not $\infty$. Here, $9x^2$ is constant with respect to $n$. – 2017-01-11

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The last line isn't correct: you should get $$ \lim_{n\to\infty}[9^nx^{2n}]^{\frac{1}{n}}=\lim_{n\to\infty}9x^2=9x^2 $$ If this limit is less than $1$ then the series converges absolutely, while if the limit is greater than $1$ then the series diverges. Hence we need $$ 9x^2<1 $$ or $|x|<\frac{1}{3}$, so $R=\frac{1}{3}$.

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    Regarding the above correct answer, always importance with these questions to ensure you don't confuse x's and n's, especially when it isn't as nice as being x-squared, so that the sign might matter. – 2017-01-11
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    Ohh yes I see. Quite an obvious mistake. Thank you! But now I'm confused how to determine my radius. So far it always was $0$ or $\infty$. But I think now you need to put more thought into it, right? – 2017-01-11
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    Yes, when using the root test or ratio test to find the radius of convergence, take the resulting limit and determine the values of $x$ for which it is less than 1. The endpoints have to be dealt with separately. – 2017-01-11
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    @carmichael561 How would that turn out in this case? – 2017-01-11
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    You mean the endpoints? Just plug them in and use some other test. In this case the series diverges at both $x=\frac{1}{3}$ and $x=-\frac{1}{3}$, using the limit test. – 2017-01-11
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    @carmichael561 I guess I have to go to sleep and look at it tomorrow again. Thanks a lot anyways. I guess I'm just too tired to realise easy things. – 2017-01-11