Find the range of $x$ for the convergence of the series $\sum_{n=1}^{\infty} \frac{1}{\left(-3\right)^{n+2}} \frac{\left(4x-12\right)^{n}}{n^{2}+1}$ 
Question: Find the range of $x$ for the convergence of the series$$\sum_{n=1}^{\infty} \frac{1}{\left(-3\right)^{n+2}} \frac{\left(4x-12\right)^{n}}{n^{2}+1}$$
  

MY Approach: By $n$th term divergence test,
$$\lim_{n\rightarrow\infty} \frac{1}{\left(-3\right)^{n+2}} \frac{\left(4x-12\right)^{n}}{n^{2}+1}=0 \Longleftrightarrow |4x-12|<1\Longleftrightarrow\frac{11}{4}<x<\frac{13}{4},$$
not in the options.
Edit 

 A: Using Cauchy root test we have:$$\lim_{n\to\infty}|a_n|^{\frac{1}{n}}=\lim_{n\to\infty}\frac{1}{3^{1+\frac{2}{n}}}|4x-12|\frac{1}{(n^2+1)^{\frac{1}{n}}}=\frac{1}{3}|4x-12|$$then for $|4x-12|<3$ the series converges and for $|4x-12|>3$ it diverges. Also for $|4x-12|=3$ we have:$$\sum_{n=1}^{\infty}\frac{1}{\left(-3\right)^{n+2}}\frac{\left(4x-12\right)^{n}}{n^{2}+1}=\sum_{n=1}^{\infty}\frac{1}{\left(-3\right)^{n+2}}\frac{1}{n^{2}+1}$$or$$\sum_{n=1}^{\infty}\frac{1}{\left(-3\right)^{n+2}}\frac{\left(4x-12\right)^{n}}{n^{2}+1}=\sum_{n=1}^{\infty}\frac{1}{3^{n+2}}\frac{1}{n^{2}+1}$$which converges in both cases. Then the domain is $|4x-12|\le 3\to2.25\le x\le 3.75$
A: Apply the root test to the above power series. A careful examination gives option $(D)$.
A: $$\sum_{n=1}^{\infty}\frac{1}{\left(-3\right)^{n+2}}\frac{\left(4x-12\right)^{n}}{n^{2}+1}=\sum_{n=1}^{\infty}\frac{1}{\left(9\right)}\frac{\left(4-\frac{4x}{3}\right)^{n}}{n^{2}+1}$$
according to Cauchy root test , we get
$$|4-\frac{4x}{3}|\leqslant 1$$
so
$$\frac{9}{4}\leqslant x\leqslant \frac{15}{4}$$
