Question

Find the radius and interval of convergence of the power series ΣΗ where a > 1. nl

Answer 1

Implies Sigma^infinity_n = 1 n!/a^n^2 x^n where a greaterthan 1 given Compare this power series Sigma a_nx^n we get a_n = n!/a^n^2 n!/a^n^2 Now a_n + 1/a_n = (n + 1)!/a^(n + 1)^2 a^n^2/n! = (n + 1)/a^2n + 1 = u_n (say) Now lim_n Rightarrowinfinity u_n + 1/u_n = Lim_n Rightarrowinfinity n + 2/a^2n + 3 a^2n + 1/n + 1 = 1/a^2 < 1 as a greaterthan 1 So lim_n Rightarrowinfinity mode a_n + 1/a_n = 0 Radius of convergence of the given power series is infinity And the interval of convergence of the power series is (- infinity infinity)