Convergent Series Homework: For What Values of p Does It Converge?

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Homework Help Overview

The discussion revolves around the convergence of the alternating series given by [1/1^p - 1/2^p + 1/3^p - 1/4^p +...], specifically focusing on determining the values of p for which the series converges.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the conditions for convergence of alternating series, referencing the need for the terms to decrease in absolute value and converge to zero. There is uncertainty about how to demonstrate convergence to zero and how to find an appropriate N for the epsilon definition.

Discussion Status

Some participants have provided insights into the requirements for convergence, including the alternating series test. There is ongoing exploration of specific values of p and their implications for convergence, with no explicit consensus reached yet.

Contextual Notes

Participants are considering the case where p is a non-negative real number and are grappling with the implications of this assumption on convergence.

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Homework Statement



For what values of p does the series [1/1^p - 1/2^p + 1/3^p - 1/4^p +... converge?

Homework Equations





The Attempt at a Solution



I believe that this series converges for all p \in N because the sequence of a_n's is nonincreasing and converges to 0. I am not quite sure, however, to show that it converges to 0. I know that the sequence 1/n converges to 0 and I know that p is fixed, but I don't know how to massage that information into what I need. Thanks.
 
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If the series is alternating, then you only need to show that [tex]|a_n| \leq |a_{n-1}|[/tex] and that the sequence of ans converges to zero.

A sequence converges to zero if for any positive real number [tex]\epsilon[/tex], you can find a N large enough so that [tex]a_n < \epsilon[/tex] for all n>N.
 
Right; I understand the epsilon proof and the theorem related to alternating series. Although I know it sounds really dumb, I am having trouble finding N.
 
[tex]\frac{1}{N^p} = \epsilon.[/tex]

Now, for certain kinds of p you can always find an N for every epsilon.
 
I was thinking for p >= 0. Thank you so much.
 

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