Find radius of convergence and interval of converges of the series; $\displaystyle\sum_{n=0}^{\infty}\frac{1}{(-3)^{2+n}(n^2+1)}(4x-12)^n$ $\displaystyle\sum_{n=0}^{\infty}\frac{1}{(-3)^{2+n}(n^2+1)}(4x-12)^n$
My try:$$\displaystyle\lim_{n\to\infty}\left(\frac{1}{(-3)^{2+n}(n^2+1)}(4x-12)^n\right)^{\frac{1}{n}}\leq1$$ $$\displaystyle\lim_{n\to\infty}\frac{1}{(-3)^{2/n+1}(n^2+1)^{1/n}}(4x-12)\leq1$$ $$\frac{1}{(-3)\displaystyle\lim_{n\to\infty}(n^2+1)^{1/n}}(4x-12)\leq1$$
Here I found $|4x-12|\leq3$, then what is radius of convergence?
Thank you.
 A: I personally prefer the ratio test. If a series converges, then:
$$\lim_{k\to \infty}\left|\frac{a_{k+1}}{a_k}\right|<1$$
Substituting gives:
$$\lim_{k\to \infty}\left|\frac{\left(\frac{(4x-12)^{k+1}}{(-3)^{3+k}((k+1)^2+1)}\right)}{\left(\frac{(4x-12)^{k}}{(-3)^{2+k}(k^2+1)}\right)}\right|<1$$
$$\lim_{k\to \infty}\left|\frac{(4x-12)^{k+1}}{(4x-12)^k}\cdot \frac{(-3)^{2+k}}{(-3)^{3+k}}\cdot \frac{k^2+1}{(k+1)^2+1}\right|<1$$
$$\lim_{k\to \infty}\left|(4x-12)\cdot \frac{-1}{3} \cdot \frac{k^2+1}{(k+1)^2+1}\right|<1$$
Taking limits gives:
$$\left|(4x-12)\cdot \frac{-1}{3}\cdot 1\right|<1$$
$$|4x-12|<3$$

To find the radius of convergence, put it in the form $|x-a|<R$ where $R$ is the radius of convergence. Now, continuing from what you've done:
$$4|x-3|<3$$
$$|x-3|<\frac{3}{4}$$
This gives the radius of convergence $R=\frac{3}{4}$.
Therefore, we obtain:
$$\frac{9}{4}<x<\frac{15}{4} \tag{1}$$
To find the interval of convergence, test the lower and upper bounds of the inequality by substituting them into your sum (i.e. $x=\frac{9}{4}$ and $x=\frac{15}{4}$) and check if they converge. If one of these bounds converge, put the $\leq$ sign next to that bound. Otherwise, leave it as $<$.
