You can use the root test here. The series will converge if
\(L=\displaystyle\lim_{n\to\infty}\sqrt[n]{\frac{(4-x)^n}{4^n+9^n}}<1\)
You have
\(L=\displaystyle\lim_{n\to\infty}\sqrt[n]{\frac{(4-x)^n}{4^n+9^n}}=|4-x|\lim_{n\to\infty}\frac1{\sqrt[n]{4^n+9^n}}\)
Notice that
\(\dfrac1{\sqrt[n]{4^n+9^n}}=\dfrac1{\sqrt[n]{9^n}\sqrt[n]{1+\left(\frac49\right)^n}}=\dfrac1{9\sqrt[n]{1+\left(\frac49\right)^n}}\)
so as \(n\to\infty\), you have \(\left(\dfrac49\right)^n\to0\), which means you end up with
\(L=\dfrac{|4x-1|}9<1\implies |4x-1|<9\implies-2
This is the interval of convergence. The radius of convergence can be determined by finding the half-length of the interval, or by solving the inequality in terms of \(|x-c| so that \(R\) is the ROC. You get
\(|4x-1|<9\implies\left|x-\dfrac14\right|<\dfrac94\implies R=\dfrac94\)
