Show that the Ʃ 1/(2n-1)^3 Converges

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SUMMARY

The series Ʃ 1/(2n-1)^3 converges, as established through the comparison test with the Riemann zeta function at s=3, denoted as zeta(3). The integral test can also be applied to demonstrate the convergence of the integral ∫x^(-3)dx from 1 to infinity, confirming that it is finite. The p-series test for p=3 provides a straightforward method for establishing convergence without needing the integral test. The discussion emphasizes clarity in terminology, recommending the use of summation notation 1/n^3 for broader understanding.

PREREQUISITES
  • Understanding of series convergence tests, specifically the comparison test and p-series test.
  • Familiarity with the Riemann zeta function and its convergence properties.
  • Knowledge of integral calculus, particularly the evaluation of improper integrals.
  • Basic understanding of mathematical notation and summation conventions.
NEXT STEPS
  • Study the properties and applications of the Riemann zeta function, particularly zeta(3).
  • Learn the details of the p-series test and its implications for convergence.
  • Explore the integral test for convergence in greater depth, including examples.
  • Review the comparison test with various series to solidify understanding of convergence criteria.
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Mathematics students, educators, and anyone interested in series convergence, particularly those studying calculus or advanced mathematical analysis.

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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
 
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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.
 

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