# Finding the radius of convergence of $\sum_{n=0}^{\infty} 3^{n} \cdot x^{2n}$

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?

• Computational error. $L = \lim |3x^2|$. – xbh Nov 12 '18 at 3:32
• thank you i fixed it – Dillain Smith Nov 12 '18 at 3:35
• It is Cauchy-Hadamard theorem, not Abel's method. But this is an overkill here since you have a geometric series. – user10354138 Nov 12 '18 at 3:36
• i like overkill – Dillain Smith Nov 12 '18 at 3:38

Here $$a_k=3^n$$ if $$k=2n$$, otherwise $$a_k=0$$. Thus $$\frac{1}{R}=\limsup|a_k|^{1/k}=\sqrt{3}$$.
• shouldn't we get $1/R = \sqrt{3}$? – Dillain Smith Nov 12 '18 at 4:00