a) determine where f(z) is analytic.
f(z) = sum from n to inf. [ ( (-1)^n * n^2 ) / ( 2^n ) * (z - 6)^n ]
b) evaluate f'''(6)
The Attempt at a Solution
a) I used the ratio test and found that f(z) is analytic on the circle of convergence C: | z - 6 | < 1/2.
b) I found the 3rd derivative of the sum as
f'''(z) = sum from 0 to inf [ ( (-1)^n * n^2 ) / ( 2^n ) * n * (n-1) * (n-2) * (z - 6)^(n-3) ]
then when i plug in 6 into this i get
f'''(6) = 0
any suggestions or comments are much apperciated. thanks