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

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In summary, the conversation was about solving a complex series using the root test. The question arose about the second step, where the numerator becomes 1. It was clarified that the principal cube root of i has a modulus of 1, and any integer power of this number also has a modulus of 1. The calculation of |i^n| was then explained using Euler's formula.
  • #1
DottZakapa
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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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  • #2
Is this supposed to be ##\displaystyle{ i ^{n/3}}## in the numerator?
 
  • #3
fresh_42 said:
Is this supposed to be ##\displaystyle{ i ^{n/3}}## in the numerator?
yes
 
  • #4
fresh_42 said:
Is this supposed to be ##\displaystyle{ i ^{n/3}}## in the numerator?
I've fixed it...
 
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  • #5
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.
 
  • #6
DottZakapa said:
yes
Then how do you calculate ##\left| i^n\right|##?
 
  • #7
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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FAQ: Why does the numerator become 1 in the root test for solving series?

1. What is the root test for solving series?

The root test is a mathematical method used to determine the convergence or divergence of a series. It is based on taking the nth root of the absolute value of the terms in the series and checking for convergence or divergence of the resulting sequence.

2. Why does the numerator become 1 in the root test?

The numerator in the root test becomes 1 because we are taking the nth root of the absolute value of the terms in the series. This is a simplification step that allows us to easily determine the convergence or divergence of the series.

3. How does the root test work?

The root test works by taking the nth root of the absolute value of the terms in the series. If the resulting sequence converges to a value less than 1, then the series is convergent. If the resulting sequence diverges or converges to a value greater than 1, then the series is divergent.

4. When should I use the root test?

The root test is most commonly used when the terms in the series involve powers or roots. It can also be used to test for convergence or divergence of alternating series.

5. What is the significance of the root test?

The root test is an important tool in determining the convergence or divergence of series. It allows us to quickly and easily determine the behavior of a series without having to use more complex methods. It is also useful in proving the convergence or divergence of a series mathematically.

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