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

1 Answer
Feb 20, 2017

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#