What is the procedure for testing convergence of alternating series?

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    Convergence Testing
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Discussion Overview

The discussion revolves around the procedure for testing the convergence of alternating series, specifically addressing two examples provided by a participant. The scope includes theoretical understanding and application of convergence tests for series.

Discussion Character

  • Homework-related
  • Technical explanation
  • Conceptual clarification

Main Points Raised

  • One participant expresses uncertainty about the procedure for testing convergence of alternating series and presents two specific series for analysis.
  • Another participant explains that an alternating series converges if the absolute value of its terms approaches zero, providing a conceptual analogy.
  • For the first series, it is noted that cos(n * pi) alternates between 1 and -1, leading to a series that can be analyzed for convergence by checking the limit of the terms as n approaches infinity.
  • Participants discuss the need to show that the limit of the terms in the first series approaches zero as n increases.
  • For the second series, it is suggested to rewrite it in a form that highlights its alternating nature and check the limit similarly.
  • One participant confirms their calculations for both series, stating that they found the limits to be zero, suggesting convergence based on their understanding of the alternating series theorem.
  • Another participant agrees with the conclusion reached regarding the convergence of the series, emphasizing the simplicity of the analysis despite initial concerns.

Areas of Agreement / Disagreement

Participants generally agree on the procedure for testing convergence of the series presented, with some confirming the steps and reasoning involved. However, there is no explicit consensus on the correctness of the final conclusions regarding convergence, as the discussion does not delve into further verification of the calculations.

Contextual Notes

Some assumptions about the behavior of the series and the application of the alternating series test are made, but these are not fully explored or verified within the discussion.

noboost4you
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We just learned Alternating Series, but I am still unsure of the procedure. A couple of problems in the book have stumped me, and I need to test them for convergence.

1. Sum n=1 going to infinity of [cos(n*pi) / (n^(7/8))]
2. Sum n=1 going to infinity of (-n/5)^n

I am really lost, I'd really appreciate the help...
 
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Okay, an alternating series converges if the absolute value of the terms in it approaches zero. It's basically as if you take a big step forwards, a smaller step back, an even smaller step forwards and so on. Try it yourself (but not in class or you'll get funny looks).

1) cos(n * pi) is 1 for n even and -1 for n odd. So your series is basically just

-1/(1^7/8) + 1/(2^7/8) -1/(3^7/8) + ...

All you have to do now is show that as n->infinity, 1/(n^7/8) -> 0, and you should have no problem with that.

2) Just rewrite this as ((-1)^n)/(5^n) and observe that (-1)^n is positive for even n and negative for odd n, so you have an alternating series. Then this question is pretty much the same as the last one.
 
so for #1, i should check the lim as n -> infinity of bn and see if it equals 0? and bn in this case is [cos(n*pi) / (n^(7/8))] ... cause according to the alternating series theorem, if the limit = 0, then it is convergent...

havent gotten to #2 yet. stay tuned ;)
 
Yeah, that's pretty much right. For these questions you need to a) show they're alternating and b) the limit as n -> inf is zero.
 
ok, did them, let's see if they're right.

1) lim n->inf [cos(n*pi) / (n^(7/8))] = 1/inf == 0 therefore, the Sum is convergent

2) lim n->inf 5^(-n) == 0 and therefore, the Sum is also convergent

yes? no? thanks for your help
 
Bingo.

Wasn't as hard as all that cos(n*pi) stuff made it look, eh? Just be glad they didn't ask you what the thing actully converged to; that's much harder.
 
oh i know, thanks again for the help. really helped a lot. I'm always pleased with the helpfulness this site brings.
 

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