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

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Answer 1 · 2016-03-06T08:57:40

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 $AA x in (-8/2, 8/2).$

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.

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

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

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

What is the radius of convergence of sum_1^oo ((2x)^n ) / 8^nsum_1^oo ((2x)^n ) / 8^n?