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Limits of sequences

  1. Dec 14, 2012 #1
    In this link:

    http://people.ischool.berkeley.edu/~johnsonb/Welcome_files/104/104hw3sum06.pdf [Broken]

    For number 8.4...

    Why don't we just say...

    [tex]|s_n||t_n| < \frac{\epsilon}{M} M = \epsilon[/tex]?

    Thanks in advance
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Dec 14, 2012 #2

    HallsofIvy

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    If you mean, choose N0 such that if n> N0 then [itex]|s_n|< \frac{\epsilon}{M}[/itex] rather than M+1, that would, give us [itex]|s_nt_n|= \epsilon[/itex], not "<" which is required for the definition of convergence.
     
    Last edited: Dec 14, 2012
  4. Dec 14, 2012 #3
    No I didn't say that...

    I said [tex]|s_nt_n| < \frac{\epsilon}{M} M = \epsilon [/tex]. So the first sign is an inequality.
     
  5. Dec 14, 2012 #4

    HallsofIvy

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    Yes, you are right about what you said and I have edited my post to remove "as you said". But you are incorrect that it would be "<". You would have, instead, "=", as I said.
     
  6. Dec 14, 2012 #5
    Probably because "we" didn't think that closely when "we" wrote that paper. If you ensure that ##|s_n|<\frac\epsilon M## what you actually get is
    $$
    |s_nt_n| = |s_n|\cdot|t_n| < \frac\epsilon M\cdot|t_n| \leq \frac\epsilon M\cdot M = \epsilon,
    $$
    which is what you have yourself.

    If you look at that paper again, you'll see that the author writes
    $$
    \begin{eqnarray*}
    |s_nt_n − 0| & = & |s_n| \cdot |t_n| \\
    & < & \left|\frac\epsilon{M + 1}\right| \cdot |M| \\
    & < & \epsilon.
    \end{eqnarray*}
    $$
    Why did he suddenly need absolute values in the middle line? I think he probably didn't proofread what he'd written.
     
  7. Dec 14, 2012 #6
    Thanks :)
     
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