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Do you think my answer is correct?

  1. Dec 14, 2012 #1
    Do you think my answer is correct? If not, can you tell me why it is wrong?

    Thanks in advance
     

    Attached Files:

  2. jcsd
  3. Dec 14, 2012 #2
    This looks fine, but you need to slightly re-adjust the inductive step. There you assume that ##|s_{n+k}-s_n|<\epsilon## and then carry on with
    $$
    \begin{eqnarray*}
    |s_{n+k+1}-s_n|
    & = & |s_{n+k+1}-s_{n+k}+s_{n+k}-s_n| \\
    & \leq & |s_{n+k+1}-s_{n+k}|+|s_{n+k}-s_n| \\
    & < & \frac\epsilon2 + \frac\epsilon2.
    \end{eqnarray*}
    $$
    That last line should be "##\ldots<\epsilon+\epsilon##". Which means you need to go back in your proof and somehow ensure that ##|s_{n+k}-s_n|## and ##|s_{n+k+1}-s_{n+k}|## are together less than ##\epsilon##, not each separately.
     
  4. Dec 14, 2012 #3
    Ok, thanks a lot :smile:
     
  5. Dec 31, 2012 #4
    Question 10.6 in this link:
    http://people.ischool.berkeley.edu/~johnsonb/Welcome_files/104/104hw3sum06.pdf [Broken]

    is the same as the one that I solved...

    But in part (b), it says that this result is not true for [tex]|s_{n+1} - s_n| < 1/n [/tex]. And I was a bit confused...because can't I use the same proof for this one too?

    Thanks in advance
     
    Last edited by a moderator: May 6, 2017
  6. Dec 31, 2012 #5

    pasmith

    User Avatar
    Homework Helper

    Unfortunately your present proof doesn't work. Your assumption is that [itex]|s_{n+k} - s_n| < \epsilon[/itex], and you know that [itex]|s_{n+ k +1} - s_{n+k}| < 2^{-(n+k)}[/itex]. The inductive step is then
    [tex]
    |s_{n+k + 1} - s_n| \leq |s_{n+k+1}-s_{n+k}| + |s_{n+k} - s_n| < 2^{-(n+k)} + \epsilon
    [/tex]
    But, sadly, [itex]2^{-(n+k)} + \epsilon > \epsilon[/itex] for all [itex]n[/itex] and [itex]k[/itex]. So you can't conclude that [itex]|s_{n+k+1}- s_n| < \epsilon[/itex], even though this might be the case. (If your inductive step doesn't make use of the given bound on [itex]|s_{n+k +1} - s_{n+k}|[/itex], then your proof is almost certainly flawed.)

    The best proof is probably the one given.
     
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