Difference between revisions of "Comparison test for real series/Statement"

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(Created page with "<noinclude> {{Requires references|grade=D|msg=Routine, but a reference would be good}} __TOC__ ==Statement== </noinclude>Suppose {{M|(a_n)_{n\in\mathbb{N} } }} and {{M|(b_n)_{...")
 
m (Typo)
 
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</noinclude>Suppose {{M|(a_n)_{n\in\mathbb{N} } }} and {{M|(b_n)_{n\in\mathbb{N} } }} are [[real sequences]] and that we have:
 
</noinclude>Suppose {{M|(a_n)_{n\in\mathbb{N} } }} and {{M|(b_n)_{n\in\mathbb{N} } }} are [[real sequences]] and that we have:
 
# {{M|\forall n\in\mathbb{N}[a_n\ge 0\wedge b_n\ge 0]}} - neither sequence is non-negative, and
 
# {{M|\forall n\in\mathbb{N}[a_n\ge 0\wedge b_n\ge 0]}} - neither sequence is non-negative, and
−
# {{M|\exists K\in\mathbb{N}\forall n\in\mathbb{N}[n>K\implies a_n\le b_n}} - i.e. that {{link|eventually|sequence}} {{M|a_n\le b_n}}.
+
# {{M|\exists K\in\mathbb{N}\forall n\in\mathbb{N}[n>K\implies b_n\ge a_n]}} - i.e. that {{link|eventually|sequence}} {{M|b_n\ge a_n}}.
 
Then:
 
Then:
 
* if {{M|\sum^\infty_{n\eq 1}b_n}} {{link|converges|sequence}}, so does {{M|\sum^\infty_{n\eq 1}a_n}}
 
* if {{M|\sum^\infty_{n\eq 1}b_n}} {{link|converges|sequence}}, so does {{M|\sum^\infty_{n\eq 1}a_n}}

Latest revision as of 06:16, 23 November 2016

Grade: D
This page requires references, it is on a to-do list for being expanded with them.
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.
The message provided is:
Routine, but a reference would be good

Statement

Suppose [ilmath](a_n)_{n\in\mathbb{N} } [/ilmath] and [ilmath](b_n)_{n\in\mathbb{N} } [/ilmath] are real sequences and that we have:

  1. [ilmath]\forall n\in\mathbb{N}[a_n\ge 0\wedge b_n\ge 0][/ilmath] - neither sequence is non-negative, and
  2. [ilmath]\exists K\in\mathbb{N}\forall n\in\mathbb{N}[n>K\implies b_n\ge a_n][/ilmath] - i.e. that eventually [ilmath]b_n\ge a_n[/ilmath].

Then:

  • if [ilmath]\sum^\infty_{n\eq 1}b_n[/ilmath] converges, so does [ilmath]\sum^\infty_{n\eq 1}a_n[/ilmath]
  • if [ilmath]\sum^\infty_{n\eq 1}a_n[/ilmath] diverges so does [ilmath]\sum^\infty_{n\eq 1}b_n[/ilmath]

References

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