Difference between revisions of "Comparison test for real series"

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Latest revision as of 07:28, 23 November 2016

Stub grade: A*
This page is a stub
This page is a stub, so it contains little or minimal information and is on a to-do list for being expanded.The message provided is:
Flesh out, link, then demote. This is needed for functional analysis

Statement

Suppose (an)nN and (bn)nN are real sequences and that we have:

  1. nN[an0bn0] - neither sequence is non-negative, and
  2. KNnN[n>Kbnan] - i.e. that eventually bnan.

Then:

  • if n=1bn converges, so does n=1an
  • if n=1an diverges so does n=1bn

Proof

Case 1

Grade: C
This page requires one or more proofs to be filled in, it is on a to-do list for being expanded with them.
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Routine for first years so unimportant. Forgive my messy notes

Case 2

Grade: C
This page requires one or more proofs to be filled in, it is on a to-do list for being expanded with them.
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Routine, but would be good to do

References

Grade: C
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Please note that this does not mean the content is unreliable, it just means that the author of the page doesn't have a book to hand, or remember the book to find it, which would have been a suitable reference.
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See correspondence with David Guichard on 22/11/2016 for where I sourced this