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Doing shady things with limits

  1. Jul 1, 2008 #1
    Given [itex]y_n = (2\log{2}+3\log{3}+...+n\log{n})/(n^2\log{n})[/itex], the problem is to find [itex]\lim_{n \to \infty} y_n[/itex]. I've found that

    [tex]y_n \left(n-1+\frac{1}{n+1}\right) \frac{\log{n}}{\log{(n+1)}} = y_{n+1} (n+1) - 1[/tex].

    So [itex] \lim_{n \to \infty} y_n(n-1) = \lim_{n \to \infty} y_{n+1} (n+1) - 1[/itex]. If I assume [itex]\lim_{n \to \infty} y_n \neq 0[/itex], separate those limits and solve to get [itex]\lim_{n \to \infty} y_n = \frac{1}{2}[/itex]. How do I show that [itex]\lim_{n \to \infty} y_n \neq 0[/itex]?
     
  2. jcsd
  3. Jul 1, 2008 #2

    Hurkyl

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    How about by induction?

    Sure, induction doesn't carry all the way through to the limit, but it can prove things about the individual terms of the sequence -- and then you can invoke some theorem on limits to 'extend' your result to the limit.

    (Incidentally, you've not only assumed the limit of y_n is nonzero, but that it exists....)

    I notice another problem. If [itex]\lim_{n \rightarrow \infty} y_n[/itex] exists and is positive, then [itex]\lim_{n \rightarrow \infty} y_n(n-1)[/itex] diverges to infinity.
     
  4. Jul 1, 2008 #3
    I don't see what I can induct. The sequence decreases for a while and then increases, so I can't say it's strictly increasing or anything like that, can I? Also, if [itex]\lim_{n \rightarrow \infty} y_n = L[/itex] is a non-zero real number, then [itex]1 = L \lim_{n \to \infty} (n+1) - L \lim_{n \to \infty} (n-1) = L \lim_{n \to \infty} ((n + 1) - (n - 1))[/itex]. Is that not valid?
     
  5. Jul 1, 2008 #4

    matt grime

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    What you did was not valid. You can only do things like

    lim(xy)=lim(x)lim(y)

    lim(x+y)=lim(x) +lim (y)

    if all the individual limits exist.

    However that isn't really an issue, since you could have rearranged the n+1's equation you got prior to taking limits.

    You still need to show that the limit of y_n exists. Since you know it is eventually increasing, then it suffices to show that it is bounded above (which it is by the most obvious upper bound you can use).
     
  6. Jul 1, 2008 #5

    tiny-tim

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    integrate?

    Hi durt! :smile:

    Is there some theorem you can use that involves integrating x logx?
     
  7. Jul 1, 2008 #6
    Well that works with some squeeze theoreming :smile:. Cool.
     
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