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Homework Help: Converging / Diverging - What is it?

  1. Mar 11, 2008 #1
    [SOLVED] Converging / Diverging - What is it?

  2. jcsd
  3. Mar 11, 2008 #2
    Wtf ??? LOL
  4. Mar 11, 2008 #3
    Studying "improper integrals". Apparently, when you use the limit (The "improper" part), the result tells you something about it being converging or diverging, and if converging, it "converges" at a point.

    I haven't a clue what the hell my book is talking about.
    The alcohol probably isn't helping either :)
  5. Mar 11, 2008 #4
    LOL, post your question. I haven't worked these type of problems in a long timeee. Would be nice to review some.
  6. Mar 11, 2008 #5
    I don't really have any specific questions that I need this for, it's just so I understand.

    For your own sake, one of the ones in the back is...
    [tex]\int_{-\infty}^{\infty} \frac{1}{1 + x^2} dx[/tex]
  7. Mar 11, 2008 #6
    Convergence and divergance basically revolve around limits. Are you studying integrals, or series and sequences? With improper integrals(which I think you are studying), all you need to to is find the limit of the eqaution; if it does dot approach a specific number as n approaches infinity it diverges, and if it does, then it converges. Consider the equation: Itegral (from 1 to infinity) of ((1/x) dx). Taking the integral of this gives you the lim as t approaches infinity of (ln(x)|(from 1 to t). (t=infinity). This equals the lim as t approaches infinity of (ln(t) - ln(1)) which = ln(t). If increasing numbers are placed into the ln() the overall number will increase until infinity is reached, meaning the equation is divergent (no limit). Im sorry i cannot make these equations out in paper written form (im new to this internet forum stuff). Hope it helps!
  8. Mar 12, 2008 #7
    Figured it out. In case anyone else needs the answer, when you take the limit, if it goes to infinity or otherwise does not exist, it diverges.

    If it does not, the limit will reduce to some sort of term. In this case, the function "converges" at that term.
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