In General

For (-1)^n cn (x-a)^n

The ratio test yields |x-a|lim{c(n+1)/cn] < 1 where (n+1) is a subscript

|x-a| < lm{cn/c(n+1}

lm{cn/c(n+1} = R is called the radius of convergence

This guarantees convergence on the interval

|x-a| < R 0r (a-R < x < a+R)

Then check x =a-R and a +R in the original series to finish the interval of convergence

In your case cn = 1/3^(2n) ????? or 1/9n ?

If 1/3^(2n) Then R = 9 you have convergence on (-7,11) check the endpoints