When representing a function as a power series will the interval of convergence always be on an open interval

f(x)= 2/(3-x) so i found the power series and got E (2x^n)/3^(n+1)
then i solved for the radius of convergence and got |x|< 3 , so to find the interval of convergence i tested x=3 and x=-3 , and i got convergent and divergent respectively. so it should be (-3,3].
the answer is supposed to be (-3,3), but i'm almost 100% sure i did this right.
could someone help in letting me know if i did it wrong or if the interval of convergence is always an open interval when finding the power series from a function, Thank you for your time.

When x = −3, then power series becomes: 2/3 − 2/3 + 2/3 − 2/3 + 2/3 ....
which is divergent (sum switches infinitely from 0 to 2/3)

When x = 3, then power series becomes: 2/3 + 2/3 + 2/3 + 2/3 + 2/3 ....
which is also divergent (it diverges to infinity)

Also, if you look at original function, f(3) = 2/0 ----> diverges to infinity

So clearly in this case, interval if convergence is open interval (−3, 3)

Exactly.

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