Question

Find the radius of convergence and the internval of convergence:

Answer 1

You can use the root test here. The series will converge if


`L=\displaystyle\lim_(n\to\infty)\sqrt[n]{((4-x)^n)/(4^n+9^n)}<1`

You have


`L=\displaystyle\lim_(n\to\infty)\sqrt[n]{((4-x)^n)/(4^n+9^n)}=|4-x|\lim_(n\to\infty)\frac1{\sqrt[n]{4^n+9^n}}`

Notice that


`\frac1{\sqrt[n]{4^n+9^n}}=\frac1{\sqrt[n]{9^n}\sqrt[n]{1+\left(\frac49\right)^n}}=\frac1{9\sqrt[n]{1+\left(\frac49\right)^n}}`

so as
`n\to\infty`, you have
`\left(\frac49\right)^n\to0`, which means you end up with


`L=\frac4x-19<1\implies |4x-1|<9\implies-2<x<\frac52`

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|<R` so that
`R` is the ROC. You get


`|4x-1|<9\implies\left|x-\frac14\right|<\frac94\implies R=\frac94`