<s>[INST] 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 

 [/INST] $$\sum_{n=1}^{\infty}\frac{1}{\left(-3\right)^{n+2}}\frac{\left(4x-12\right)^{n}}{n^{2}+1}=\sum_{n=1}^{\infty}\frac{1}{\left(9\right)}\frac{\left(4-\frac{4x}{3}\right)^{n}}{n^{2}+1}$$
according to Cauchy root test , we get
$$|4-\frac{4x}{3}|\leqslant 1$$
so
$$\frac{9}{4}\leqslant x\leqslant \frac{15}{4}$$
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