Question

Find the radius of convergence and the interval of convergence of the series\(\sum\limits_{n = 1}^\infty {{n^n}} {x^n}\).

Short answer

The radius of convergence is\(R = 0\) and the interval of convergence is\({\rm{I = (0,0)}}\).

Step-by-step solution

The root test.

“If\(\mathop {\lim }\limits_{n \to \infty } \sqrt(n){{{a_n}}} = L < 1{\rm{ or }}\mathop {\lim }\limits_{n \to \infty } \sqrt(n){{\left| {{a_n}} \right|}} = \infty {\rm{, }}\)then the series\(\sum\limits_{n = 1}^\infty {{a_n}} \)is divergent.”

Use the root test for calculation.

The series is\(\sum\limits_{n = 1}^\infty {{n^n}} {x^n}\).

Let\({a_n} = {n^n}{x^n}\)

Then,

\(\begin{aligned}\mathop {\lim }\limits_{n \to \infty } \sqrt(n){{\left| {{a_n}} \right|}} &= \mathop {\lim }\limits_{n \to \infty } n|x|\\ &= \infty |x|\\ &= \infty \;\;\;{\rm{ (if }}x \ne 0)\end{aligned}\)

Hence, the series diverges.

Therefore, the radius of convergence is\(R = 0\) and the interval of convergence is\({\rm{I = (0,0)}}\).