Question

Consider the geometric series S(x)=1+2(x−3)+4(x−3)^2+8(x−3)^3+⋯

Giving your answer as an interval, find all values of x for which the series converges.
Now assuming that x is in your interval above, find a simple formula for S(x)

Answer 1

Already did one like this. It converges if the common ratio satisfies |r| < 1

So, you need |2(x-3)| < 1
Then S(x) = a/(1-r) = 1/(1 - 2(x-3))