Answer:
a=1 converges for 0<x<8
Explanation:
Given that a geometric series infinite is given in summation form as
![\Sigma _0^\infty ((x-4)/(4) )^n](https://img.qammunity.org/2020/formulas/mathematics/college/jbknebtvzo05z9cdokl7qb4bogpxub4sew.png)
Here I term is when n =0
So a= I term =
![((x-4)/(4) )^0=1](https://img.qammunity.org/2020/formulas/mathematics/college/b1uoxkagfgiuc5cxymotpp22apbc1zw057.png)
r = common ratio =
![(x-4)/(4)](https://img.qammunity.org/2020/formulas/mathematics/college/k5bpi191wevyaa9q8wpiibp785vrcuwbmu.png)
This series converges only if
![|r|<1](https://img.qammunity.org/2020/formulas/mathematics/middle-school/t7yt0d8ehyo8n535gx6sqqab3fv95b13wt.png)
![|(x-4)/(4)|<1\\-4<x-4<4\\0<x<8](https://img.qammunity.org/2020/formulas/mathematics/college/dy5i7358c9a0iy09gi8ts3bqf8ksqvbw3g.png)
For 0<x<8, we get absolute value of r is less than 1.
Hence series converges for
![0<x<8](https://img.qammunity.org/2020/formulas/mathematics/high-school/fjkrv7jdyggryp429b5mr24dy4e9ck0hew.png)