Question

Please answer the question fully! Show all your work and write neatly. If you don’t know how to answer it then do not answer! If you copy someone else’s work I will downvote and report.

a) Find the radius of convergence of the power series Σ000 sin(ntor".

Answer 1

"Find the radius of convergence of the power series sigma^infinity_n = 0 sin (n pi/6)/2^n x^n sigma^infinity_n = 0 sin (n pi/6)/2n x^n = sigma^infinity_n = 0 a_n x^n sin x = e^ix - e^ix/2i a_n = sin (npi/6)/2^n = e^n pi/6 - e^-n pi/6/2i times 2^n a_n + 1 = sin ((n + 1) pi/6)/2^n + 1 = e^in pi/6 + I pi/6 - e^-ln + sin 6/2i time 2^n + 1 center of convergence is c = 0 a_n + 1/a_n = (e^i(n + 1) pi/6 - e^-I (n + 1) I pi/6)/2i times 2^n times 2 [e^i(n)pi/4 - e^-in pi/6]/2i times 2^n = e^i(n + 1) pi/6 - e^I (n + 1) pi/6/2[e^in pi/6 - e^-in pi/6] = 1/2 e^i(n + 1)pi/6/e^in pi/6 [1 - e^2i n + D pi/6]/[1 - e^2 i(n pi/6)] since a_n + 1/a_n = 1/2 e^-I pi/6 [1 - e^-2i (n + 1) n/6/1 - e^-2i lnm pi /6] L = lim_n rightarrow infinity