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Proving a limit and the Order Limit Theorem

  1. Feb 13, 2008 #1
    If [tex](a_n) \rightarrow 0[/tex] and [tex]\left|b_n - b\right| \leq a_n[/tex], then [tex](b_n) \rightarrow b[/tex].

    I'm a little bit stuck on proving this! Part (iii) of the Order Limit Theorem states:

    Assume:[tex]lim a_n = a[/tex] and [tex]lim b_n = b[/tex]
    If there exists [tex]c \in[/tex] R for which [tex]c \leq b_n[/tex] for all [tex]n \in[/tex] N, then [tex]c \leq b[/tex]. Similarly if [tex]a_n \leq c[/tex] for all [tex]n \in[/tex] N, then [tex]a \leq c[/tex].

    I can then take this to a point where [tex]b_n = b[/tex], but somehow that doesn't seem right because [tex]\left|b_n - b\right|<\epsilon[/tex], where [tex]\epsilon > 0[/tex]. Where am I going wrong here!
     
  2. jcsd
  3. Feb 13, 2008 #2
    What happens if you choose n so large that |an| is less than epsilon? Why can you choose such an n?
     
  4. Feb 13, 2008 #3
    That's exactly what I want. I want |an| to be less than epsilon because it implies convergence for any n, which means |an| is practically nothing because it approaches 0.

    I'm not really sure what you are getting at here.
     
  5. Feb 13, 2008 #4
    You know {an} converges to 0, what does that tell you?
     
  6. Feb 13, 2008 #5
    [tex]b_n - b = 0[/tex]

    So, basically, [tex]b_n = b[/tex]

    But I'm wondering if that's really all thats necessary?
     
  7. Feb 13, 2008 #6
    I don't understand how you get that from what I wrote.

    Now IGNORE {bn} for a minute you know that {an} converges to zero, what does that mean?
     
  8. Feb 13, 2008 #7
    We had [tex]a_n[/tex] in the inequality, that's what I thought you were referring to.

    Basically, the sequence approaches zero! Convergent series are bounded above and below, and a sequence converges when, for every positive integer [tex]\epsilon[/tex], there exists and [tex]N \in [/tex] N such that whenever [tex]n \leq[/tex] N, it follows that |an - a| < [tex]\epsilon[/tex].

    I know the definition of convergence, in this case: |an| < [tex]\epsilon[/tex]

    Shall I list the properties of convergent sequences while I'm at it too! LOL!
     
  9. Feb 13, 2008 #8
    If you are trying to prove just this part, than here you go:
    since [tex](a_n) \rightarrow 0[/tex] it means that for every e>o there exists some N(e)>0 such that whenever n>N we have Ia_nI<e, which actually is

    -e<a_n<e
    Now going back to [tex]\left|b_n - b\right| \leq a_n<e[/tex] which actually means that b_n converges to b.

    what else are u looking for?
    P.S. This is what d_leet also suggested, i guess!
     
  10. Feb 13, 2008 #9
    Two things. One please don't post entire solutions to questions such as this. And two, there actually needs to be an additional restriction to {an} or to the inequality of |b{sub]n[/sub]-b|<an in order for it to make sense; that restriction being either {an} is a sequence of positive terms or the inequality should be |bn-b|<|an|
     
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