Show that the Ʃ 1/(2n-1)^3 Converges

  • Thread starter Thread starter McAfee
  • Start date Start date
Click For Summary

Homework Help Overview

The discussion revolves around demonstrating the convergence of the series Ʃ 1/(2n-1)^3, with participants exploring various methods and tests related to series convergence.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the use of the ratio test, comparison test, and integral test to establish convergence. There is a question regarding the summation limits and whether it includes all positive integers. Some suggest comparing the series to known convergent series like the p-series or zeta(3).

Discussion Status

Multiple approaches are being explored, including the comparison to zeta(3) and the p-series test. Participants are clarifying assumptions about the series and discussing the clarity of terminology used in the comparison.

Contextual Notes

There is an assumption that the summation is over all positive integers starting from n=1. Participants are also considering the implications of using different convergence tests and their familiarity with specific mathematical concepts.

McAfee
Messages
96
Reaction score
1

Homework Statement



Show that the Ʃ 1/(2n-1)^3 Converges


The Attempt at a Solution



I tried using the ratio the ratio test but that didn't work
 
Physics news on Phys.org
I assume the summation is over all non-negative n (i.e. 1,2,3..)?

Just compare with zeta(3) (which has all positive terms and includes all the terms in your series), which converges. If you need to establish convergence of the latter, use the integral test, i.e. prove that ∫x^(-n)dx for the bounds [1, infinity) is finite for all n > 1 (here the n is 3).

EDIT: Damn Latex
 
Last edited:
Curious3141 said:
I assume the summation is over all non-negative n (i.e. 1,2,3..)?

Just use the comparison test with zeta(3), which converges. If you need to establish convergence of the latter, use the integral test.

Good advice, but it's probably clearer if you say summation 1/n^3 instead of zeta(3). Not everybody knows the Riemann zeta function.
 
Dick said:
Good advice, but it's probably clearer if you say summation 1/n^3 instead of zeta(3). Not everybody knows the Riemann zeta function.

Fair enough.:smile:
 
Dick said:
Good advice, but it's probably clearer if you say summation 1/n^3 instead of zeta(3). Not everybody knows the Riemann zeta function.

If I say 1/n^3 could I also use the p-series test.

and yes i meant the summation where n=1 to infinity
 
McAfee said:
If I say 1/n^3 could I also use the p-series test.

and yes i meant the summation where n=1 to infinity

If you're allowed to assume the p-series convergence for p = 3, that's essentially the same thing as assuming the convergence for zeta(3). Then you don't even need the integral test. Just compare and you're done.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
Replies
14
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K