Question

How do you find the radius of convergence of the power series `Sigma 2^n n^3 x^n` from `n=[0,oo)`?

Answer 1

The series:

`sum_(n=0)^oo 2^n n^3x^n`

has radius of convergence `R=1/2`

Explanation:

Given the series:

`sum_(n=0)^oo 2^n n^3x^n`

apply the ratio test to determine the radius of convergence.
Evaluate:

`lim_(n->oo) abs ( (2^(n+1) (n+1)^3 x^(n+1))/ (2^n n^3 x^n)) = lim_(n->oo) 2 ((n+1)/n)^3 abs x = 2 abs(x)`

So the series is absolutely convergent for:

`2absx < 1`

that is:

`abs x < 1/2`