internal of convergence of a power series

Question

Determine the interval of convergence of the power series which is the sum from n=0 to infinity

of the terms a_n=[n(x+2)ⁿ]/(3ⁿ⁺¹)

Answer 1

Use the Ratio Test

l a_n+1 / a_n l = l ((n+1)(x+2)ⁿ⁺¹ / 3ⁿ⁺²) (3ⁿ⁺¹ / (n(x+2)ⁿ)) = [(n+1) / n](1/3) l x+2 l

Taking the limit as n goes to infinity, we get l x + 2 l / 3

The series converges when l x + 2 l / 3 < 1

-3 < x + 2 < 3

-5 < x < 1

Check endpoints:

When x = 1, a_n = n/3 and the series diverges

When x = -5, a_n = (-1)ⁿ(n / 3) and the series diverges

Interval of convergence is (-5, 1)