Hello Yumi (that's a nice name, by the way)! Let me help you get your question answered!
You may already be familiar that there is a similar geometrical series function f(u) = 1 / ( 1 - (u) ) centered at x = 0, which has a power series: 1 + u + u ^ 2 + u ^ 3 + .... u ^ n (where 0 ≤ n ≤ ∞).
The amazing thing about series is that you can substitute numbers to create new series.
To help you understand what I mean, let u = 9x (do you see where I'm heading now?).
Go back to the series above and directly substitute every u for 9x:
1 + 9x + (9x) ^ 2 + (9x) ^ 3 + .... (9x) ^ n (where 0 ≤ n ≤ ∞) <<<<< (ANSWER to first part)
The interval of convergence can be found easily here. We know that geometric series must have |u| < 1, which means that |9x| < 1 ⇒ -1/9 < x < 1/9 >>>>> (ANSWER)
Lastly, to find the sum, we can use the fact that geometric series have a sum where: S = a / (1 - u).
In here, a = 1 (the first number in the series) and u = 9x. So the sum is literally just S = 1 / (1 - 9x). As long as x is -1/9 < x < 1/9 (yep, the interval of convergence), then any sum can be found < (ANSWER)
Hope this helped you! Please let me know if you have any more questions for me!