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Converge Diverges

  1. May 20, 2010 #1
    I would really appreciate if someone would explain Converges and Diverges to me since I do not understand them.

    For example I do not know what to do in this problem:
    (5n4 +1)/(150,348n3 +999)
    or problems like this
    2,1, (2/3), (1/2), (2/5)

    Please help! I have a test tomorrow on this!
    Last edited: May 20, 2010
  2. jcsd
  3. May 20, 2010 #2


    Staff: Mentor

    I'm assuming you're talking about convergence or divergence of a sequence, so I'll give rough definitions in that context. A sequence of numbers converges to a limit L if the terms in the sequence get arbitrarily close to L as n gets larger.

    The sequence diverges if its terms get larger and larger without bound, or they get more and more negative, or if they never settle on a particular value.

    For example, the sequence {1/n} = {1, 1/2, 1/3, 1/4, ..., 1/n, ...} converges to 0. The larger n gets, the closer 1/n gets to zero.

    The sequence {(-1)^n} = {-1, 1, -1, 1, ...} diverges.
    The sequence {n^2/(n + 500} diverges. The first few terms in this sequence are {1/501, 4/502, 9/503, 16/504,...}

    The sequence {2, 1, 2/3, 1/2, 2/5, ...} can also be written as {2/1, 2/2, 2/3, 2/4, 2/5, ...} There are other possibilities, but I would guess that the next term in the sequence is 2/6 = 1/3.
    Last edited: May 20, 2010
  4. May 20, 2010 #3


    User Avatar
    Science Advisor
    Homework Helper

    For fractions, you can ignore all but the leading terms when you want to know whether it converges or diverges. So (3x^2 + 2x - 1)/(4x^3 - 3x) converges because 3x^2 / 4x^3 converges.
  5. May 22, 2010 #4
    For the first sequence, note that the quadratic terms increase much faster than non-quadratics, and further you can show that terms of higher powers increase much faster than terms of lower powers. This notion of "higher degree terms dominate the convergence behavior" can be made more precise.
    For example, it is easy to show that the sequence (1/n) approaches 0 as n increases without bound. This means that (n/n^2) also does this, and so forth.
    It is also easy to see that (n) diverges, and thus so does (n^2/n) and so forth.
    (a*n^2)/(b*n^2) and similar forms are obviously just a/b.
    Consider (an^2 + bn + c)/(dn^2). Division shows this sequence must be a/d by the simple theorems above.
    Now for a full rational sequence: (an^3 + bn^2 + cn + d)/(en^3 + fn^2 + gn + h). Multiply the top and bottom by 1/n, and examine the behavior of each term. Keep doing so until you get to the intuitive conclusion that this sequence converges to a/e.
    You can now prove yourself that sequences of rational terms converge to 0 when the denominator is of a higher degree, converge to the ratio of the coefficients of the two highest degree terms when the degree of the numerator and denominator are equal, and diverge when the numerator is of higher degree than the denominator.
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