Consider the power series (n=1 to infinity) [tex]\Sigma[/tex] (x^n)/(n*3^n). (a) Find the radius of convergence for this series. (b) For which values of x does the series converge? (include the discussion of the end points). (c) If f(x) denotes the sum of the series, find f'(x) as explicitly as possible. I know how to do parts a and b and for the radius I got 3 and the interval looks like this: -3≤x<3 But I don't understand the last part (c) I know that since (n=0 to infinity)f(x)= [tex]\Sigma[/tex] c_n(x-a)^n then (n=1 to infinity)f'(x)=[tex]\Sigma[/tex] n*c_n(x-a)^(n-1) I was looking thru various materials and first I was thinking about finding the representation for the power series but I was not sure how this would help me here since I don't have a function to begin with just the series. Then I was thinking about differentiating by terms but I didn't know how to incorporate the radius of convergence. Finally I did something that I pretty sure is wrong: (n=1 to infinity)f'(x)=[tex]\Sigma[/tex] n*c_n(x-a)^(n-1) = [tex]\Sigma[/tex] n*(1/(n*3^n))*(x^(n-1))=(1/(3^n))*(x^(n-1))=(x^n)/(x*3^n) for -3≤x<3 But like I said it doesn't seem correct. Any suggestions?