Alternating Series and P-series convergence

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Homework Help Overview

The discussion revolves around the convergence of an alternating series defined by the expression \(\sum^{\infty}_{n=1}\frac{-1^{n-1}}{\sqrt{n}}\). Participants are exploring the application of the Alternating Series Test and the p-series test in determining the behavior of this series.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the structure of the series and the role of the terms \(b(n) = \frac{1}{\sqrt{n}}\) in the context of the Alternating Series Test. There are questions regarding the validity of applying the p-series test, with some suggesting that it may not be appropriate for this series due to its alternating nature.

Discussion Status

There is an ongoing exploration of the convergence properties of the series, with some participants suggesting a focus on the Alternating Series Test while others express confusion about the application of the p-series test. The discussion indicates that the series converges conditionally, but there is no explicit consensus on the reasoning behind this conclusion.

Contextual Notes

Participants note that the p-series test typically applies to positive series and question the assumptions regarding the convergence of alternating series. There is also mention of the distinction between absolute and conditional convergence in the context of the tests being discussed.

kylera
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Alternating Series and P-series "convergence"

I couldn't resist trying out a pun. Anyway, onto the question:

Homework Statement


Test the series for convergence/divergence:
\sum^{\infty}_{n=1}\frac{-1^{n-1}}{\sqrt{n}}


Homework Equations


Alternating Series Test and possibly p-series test...


The Attempt at a Solution


The expression for an alternating series goes as a(n) = (-1)^(n-1) * b(n). Having said that, it's obvious that b(n) is

\frac{1}{\sqrt{n}}

which can be re-written as \frac{1}{n^\frac{1}{2}}. But by the rules of the p-series, since 0.5 is obviously lesser than 1, I came to the conclusion that the series diverges. Instead, the answer states that the series does indeed converge. Can anyone help shed some light in this?
 
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I think for the p-series test, the numerator must be '1' only.
 


Forget the p-series test. Concentrate on the alternating series test. b(n) is decreasing towards 0. Sorry, I don't get the 'pun'. Am I being thick?
 


Dick said:
Sorry, I don't get the 'pun'. Am I being thick?

I don't think you are since I don't get it either.
 


Well usually for the p-series test, the only examples with fractions I've ever seen is where the function is in the form

\frac{1}{f(n)}
 


rock.freak667 said:
Well usually for the p-series test, the only examples with fractions I've ever seen is where the function is in the form

\frac{1}{f(n)}

Sure, in fact, it applies only to series of the form 1/n^p or n^(p). I wasn't saying you were wrong. It's useful here if you want to discuss absolute convergence. It is useless for the given function.
 


The "p-series" only applies to positive series. That's why Dick says it is useful to discuss absolute convergence. If your series has both positive and negative terms then it may converge "conditionally". It will converge absolutely only if the series of absolute values converges.

By the "alternating series test", the series
\sum \frac{(-1)^n}{\sqrt{n}}
converges.

By the "p- series test"
\sum \left|\frac{(-1)^n}{\sqrt{n}}\right|= \sum \frac{1}{\sqrt{n}}= \sum \frac{1}{n^{1/2}}
does NOT converge and so the original series converges conditionally, not absolutely.
 

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