What is the radius of convergence of #sum_1^oo ((2x)^n ) / 8^n#?

1 Answer
Mar 6, 2016

Answer:

The radius of convergence #R# of a power series of the form #sum_(n=0)^(oo) a_n x^n# is given by #R = 1/(lim "sup"_(n->oo)root(n)(abs(a_n))).#

In your case, #R = 8/2# and #sum_(n=1)^oo ((2x)^n ) / 8^n# converges #AA x in (-8/2, 8/2).#

Explanation:

The series #sum_(n=1)^oo ((2x)^n ) / 8^n# is already written in the wished form :

#sum_(n=1)^oo ((2x)^n ) / 8^n = sum_(n=1)^oo (2/8)^n x^n = sum_(n=1)^oo a_n x^n,# #a_n = (2/8)^n.#

Let's compute #lim "sup"_(n->oo)root(n)(abs(a_n))# :

#root(n)(abs(a_n)) = root(n)(abs((2/8)^n)) = root(n)((2/8)^n) = 2/8 AA n in NN.#

Therefore, #lim "sup"_(n->oo)root(n)(abs(a_n)) = lim_(n->oo)root(n)(abs(a_n)) = 2/8 => R = 1/(2/8) = 8/2.#

Thus, #sum_(n=1)^oo ((2x)^n ) / 8^n# converges #AA x in (-8/2, 8/2).#

You should now check the endpoints of the interval.

  • If #x = -8/2# :

#sum_(n=1)^oo (2/8)^n (-8/2)^n = sum_(n=1)^oo (-1)^n# is not defined.

  • If #x = 8/2# :

#sum_(n=1)^oo (2/8)^n (8/2)^n = sum_(n=1)^oo 1# diverges.

Thus, #sum_(n=1)^oo ((2x)^n ) / 8^n# converges #AA x in (-8/2, 8/2).#