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I need help finding the radius of convergence of $\sum_{n=0}^{\infty} 3^{n} \cdot x^{2n}$.

I recognize that this is a geometric power series with first term equal to $1$ and common ration equal to $3x^{2}$. Intuitively, I know that the geometric power series must diverge if $x > 1$ because then each term will get larger and larger, but I don't know how to show this.

I think that the correct way to do this would be to use the ratio test:

$$ \begin{align*} L = \lim_{n\to\infty} \left|\frac{a_{n + 1}}{a_{n}}\right| \\[2em] = \lim_{n\to\infty} \left|\frac{3^{n + 1}\cdot x^{2n + 2}}{3^{n} \cdot x^{2n}}\right| \\[2em] = \lim_{n\to\infty} \left|3^{} x^{2}\right| \\[2em] = 3 \cdot \lim_{n\to\infty} \left|x^{2}\right|. \end{align*} $$

Now, when $L < 1$, our series converges. Equivalently, we must have $\lim_{n\to\infty} |x^{2}| < 1/3$ . When $L > 1$, our series diverges. Equivalently, we must have $\lim_{n\to\infty} |x^{2}| > 1/3$. I'm not sure how to get the radius of convergence from here. Any help would be appreciated.

Edit: continuing on: When $|x^{2}| < 1/3,$ we have $x^{2} < 1/3$ and $-x^{2} < 1/3$. So, $x < 1/\sqrt{3}$ and $x < -1/\sqrt{3}$


Edit: I saw a similar proof using $\lim \text{sup}$, and it was significantly shorter (I think it was Abel's Method)?. How would I go about solving the problem using this method?

I think by Abel's Method, we can conclude $R = \frac{1}{\lim \text{sup}_{n\to\infty} |a_{n}|^{1/n}}$.

In our case, we would have $\lim_{n\to\infty} \text{sup} |a_{n}|^{1/n} = \lim_{n\to\infty} (3^{n})^{1/2n} = \sqrt{3}$. Thus, $R = 1/\sqrt{3}$. Is this valid?

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  • $\begingroup$ Computational error. $L = \lim |3x^2|$. $\endgroup$ – xbh Nov 12 '18 at 3:32
  • $\begingroup$ thank you i fixed it $\endgroup$ – Dillain Smith Nov 12 '18 at 3:35
  • $\begingroup$ It is Cauchy-Hadamard theorem, not Abel's method. But this is an overkill here since you have a geometric series. $\endgroup$ – user10354138 Nov 12 '18 at 3:36
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    $\begingroup$ i like overkill $\endgroup$ – Dillain Smith Nov 12 '18 at 3:38
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Here $a_k=3^n$ if $k=2n$, otherwise $a_k=0$. Thus $\frac{1}{R}=\limsup|a_k|^{1/k}=\sqrt{3}$.

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  • $\begingroup$ is what i did in my original post correct? $\endgroup$ – Dillain Smith Nov 12 '18 at 3:54
  • $\begingroup$ shouldn't we get $1/R = \sqrt{3}$? $\endgroup$ – Dillain Smith Nov 12 '18 at 4:00
  • $\begingroup$ Ohh yes. I corrected it. $\endgroup$ – Anupam Nov 12 '18 at 4:19

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