Question

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

Answer 1

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).`