((x-2)^n)/(2^n)

Find the radius

Here c_n = 1/(2^n), (c_n)^(1/n) = 1/2.

The series converges when (1/2)|x-2|<1 and diverges when>1.

So the radius is 2 around x=2, i.e. interval of convergence is (0, 4).

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