Why does the numerator become 1 in the root test for solving series?

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Homework Statement
solving a complex series applying the root test
Relevant Equations
series with complex numbers
in order to solve a series, the root test is applied and I have this limit
## \lim_{n \rightarrow +\infty} \sqrt[n] {\left| {\frac {i^{\frac { n} {3}}} { \frac {2n} {3} +1}} \right| } ##

I don't understand why at the second step the numerator becomes 1, cannot recall why it becomes 1, that is:

## \lim_{n \rightarrow +\infty} \sqrt[n] { {\frac {1} { \frac {2n} {3} +1}} } ##
 
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Is this supposed to be ##\displaystyle{ i ^{n/3}}## in the numerator?
 
fresh_42 said:
Is this supposed to be ##\displaystyle{ i ^{n/3}}## in the numerator?
yes
 
fresh_42 said:
Is this supposed to be ##\displaystyle{ i ^{n/3}}## in the numerator?
I've fixed it...
 
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DottZakapa said:
Homework Statement:: solving a complex series applying the root test
Relevant Equations:: series with complex numbers

I don't understand why at the second step the numerator becomes 1, cannot recall why it becomes 1
The principal cube root of i (i.e., ##i^{1/3}##) has a modulus of 1. Any integer power of this number produces a different angle, but the modulus remains 1.
 
DottZakapa said:
yes
Then how do you calculate ##\left| i^n\right|##?
 
fresh_42 said:
Then how do you calculate ##|i^n|##?
Write ##i## as ##e^{i\pi/2}##. Then it's clear that ##\left |i^n \right | = \left | e^{i n \pi/2} \right | = 1## for any real n.
 
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RPinPA said:
Write ##i## as ##e^{i\pi/2}##. Then it's clear that ##\left |i^n \right | = \left | e^{i n \pi/2} \right | = 1## for any real n.
Would have been nice to read this from the OP.
 
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