Question

Consider the geometric series S(x)=1+3(x−3)+9(x−3)2+27(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

geometric series converge when |r| < 1
So, we need
|3(x-3)| < 1
|x-3| < 1/3
8/3 < x < 10/3
S(x) = 1/(x-1)
Check:
S(2.8) = 1 - 0.6 + 0.36 - ... = 1/1.6 = 5/8
S(3.2) = 1 + 0.6 + 0.36 + ... = 1/0.4 = 5/2