Determining convergence of series

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    Convergence Series
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Discussion Overview

The discussion revolves around determining the convergence or divergence of the series from n=1 to infinity for the expression (-1)^n / (n^3 - ln(n)). Participants explore various convergence tests, including the Alternating Series Test, Direct Comparison Test, and Limit Comparison Test, while debating the nature of convergence (conditional vs. absolute).

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant believes the series converges through the Alternating Series Test but is uncertain about whether the convergence is conditional or absolute.
  • Another participant asserts that near infinity, the term n^3 is dominant over ln(n).
  • Some participants mention using the comparison c(x) = 1/n^3 for both Direct Comparison Test (DCT) and Limit Comparison Test (LCT), noting that 1/n^3 converges by p-series.
  • There is a claim that the limit comparison test yields a limit of 1, suggesting absolute convergence.
  • Questions arise regarding the methods used to determine absolute convergence, with one participant asking for clarification on the analysis of end behavior.
  • Another participant expresses confusion about obtaining a limit of infinity when applying the Limit Comparison Test.
  • A participant provides a detailed calculation showing how they arrived at a limit of 1 using the Limit Comparison Test.

Areas of Agreement / Disagreement

Participants express differing views on the application and results of the Limit Comparison Test, with some asserting absolute convergence while others remain uncertain about their findings. There is no consensus on the overall convergence status of the series.

Contextual Notes

Participants mention inconclusive results from the Direct Comparison Test and the Limit Comparison Test, indicating potential limitations in their analyses. The discussion includes various assumptions about the behavior of the terms in the series as n approaches infinity.

calcboi
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I have a question on which test to use for series n=1 to infinity for (-1)^n / (n^3)-ln(n) in order to determine convergence/divergence. I am pretty sure I determined it converges through the Alternating Series Test(correct me if I'm wrong) but I am not sure whether it is conditional or absolute. I tried the Direct Comparison Test but it was inconclusive, and I am stuck now on what to do. I also tried Limit Comparison but the limit goes to infinity so it is also inconclusive. Can you please help?
 
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Yes you are correct.

near infinity the term n^3 is dominant over ln (n).
 
BTW, I used the comparison c(x) = 1/n^3 for DCT and LCT since we know 1/n^3 converges by p-series.
 
calcboi said:
BTW, I used the comparison c(x) = 1/n^3 for DCT and LCT since we know 1/n^3 converges by p-series.

Right, $\displaystyle\lim_{n\to \infty}\left(\frac{1}{n^3-\log n}:\frac{1}{n^3}\right)=\ldots =1\neq 0$, so the series is absolutely convergent.
 
What test did you use to determine absolute convergence? Or was that just analyzing end behavior?
 
calcboi said:
When I tried the Limit Comparison Test, I got infinity as n approaches infinity. How did you get 1?

$\displaystyle\lim_{n\to \infty}\left(\frac{1}{n^3-\log n}:\frac{1}{n^3}\right)=\lim_{n\to \infty}\frac{n^3}{n^3-\log n}=\lim_{n\to \infty}\frac{1}{1-(\log n/n^3)}=\frac{1}{1-0}=1$
 

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