Finding the domain of absolute convergence

Apr 2016
6
0
VA
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
Sep 2012
6,608
1,263
Australia
Hey HeidiHall1995.

Have you considered any inequalities like the triangle inequality?
 
Jun 2013
1,151
614
Lebanon
\(\displaystyle \lim \frac{(n+1)\left(n^3+n^{1/3}\right)}{(n+1)^3+(n+1)^{1/3}} =\infty \)

so the radius of convergence is 0