# Finding the domain of absolute convergence

#### HeidiHall1995

Find the domain of absolute convergence for:

$$\sum_{n=1}^\infty \frac{(n!)(iz+1)^n} {n^3+ \sqrt[3] {n}}$$

which test should be used in this case?

I tried the ratio test, but the terms did not cancel out (some did):

$\lim_{n\to \infty} \frac{(n+1)!(iz+1)^{n+1}} {(n+1)^3 + \sqrt[3] {n+1}} \frac {n^3+ \sqrt[3] {n}} {(n!)(iz+1)^n}$

$\lim_{n\to \infty} \frac{(n+1)(iz+1)} {(n+1)^3 + \sqrt[3] {n+1}} \frac {n^3+ \sqrt[3] {n}} {1}$

I am kind of stuck at this point. Any help to pass this point is appreciated.

#### chiro

MHF Helper
Hey HeidiHall1995.

Have you considered any inequalities like the triangle inequality?

#### Idea

$$\displaystyle \lim \frac{(n+1)\left(n^3+n^{1/3}\right)}{(n+1)^3+(n+1)^{1/3}} =\infty$$

so the radius of convergence is 0