See if series converges using the root test

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    Root Series Test
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Discussion Overview

The discussion revolves around determining the convergence of a series using the root test, specifically examining the expression involving complex numbers. Participants explore different interpretations and calculations related to the series' terms.

Discussion Character

  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant presents the expression \(((2+i)/(3-4i))^{2n}\) and expresses uncertainty about how to derive the convergence ratio of \(1/5\).
  • Another participant proposes a different series term \(a_n=\left(\frac{2+n}{3-4n}\right)^{2n}\) and calculates the limit \(C=\lim_{n\to\infty}\sqrt[n]{\left|a_n\right|}=\frac{1}{16}<1\), suggesting absolute convergence.
  • A third participant questions the change from \(i\) to \(n\) in the series, noting that the original series is complex and referencing a book that states the answer is \(1/5\).
  • A later reply confirms the calculation of the modulus of the complex fraction, stating \(\left|\frac{2+i}{3-4i}\right|^2=\frac{2^2+1^2}{3^2+4^2}=\frac{1}{5}\).

Areas of Agreement / Disagreement

Participants express differing views on the correct interpretation of the series and the calculations involved. There is no consensus on the correct approach or final answer.

Contextual Notes

There are unresolved assumptions regarding the definitions of the series terms and the treatment of complex numbers in the context of the root test.

Logan Land
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((2+i)/(3-4i))^(2n)

I know it converges with 1/5 < 0

but I don't understand how to obtain 1/5
unless somehow its like this
(2+i)^2 = (4+1)=5
(3-4i)^2 = (9+19)=25
which is 5/25 which is 1/5
but that doesn't seem like the correct way to me
 
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I am assuming here that:

$$a_n=\left(\frac{2+n}{3-4n}\right)^{2n}$$

And so:

$$C=\lim_{n\to\infty}\sqrt[n]{\left|a_n\right|}=\frac{1}{16}<1$$

And so we may conclude the series converges absolutely.
 
MarkFL said:
I am assuming here that:

$$a_n=\left(\frac{2+n}{3-4n}\right)^{2n}$$

And so:

$$C=\lim_{n\to\infty}\sqrt[n]{\left|a_n\right|}=\frac{1}{16}<1$$

And so we may conclude the series converges absolutely.
why did you turn the i's into n's?
how the book says the answer is 1/5
it is a complex series
 
Oh, okay. Then I think what you did is right:

$$\left|\frac{2+i}{3-4i}\right|^2=\frac{2^2+1^2}{3^2+4^2}=\frac{1}{5}$$
 

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