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Why is Harmonic Series Convergent

  1. Mar 10, 2012 #1
    I understand that the harmonic series, [itex]\frac{1}{x}[/itex] is divergent because:

    [itex]\int (1/x) [/itex]
    from one to infinity is:

    [ln(infinity) - ln(1)]

    which is clearly divergent.


    When I look at the graph of [itex]\frac{1}{x}[/itex] versus [itex]\frac{1}{x^{2}}[/itex]
    they both look like they are converging to zero as
    x approaches infinity.

    Whats the deal?
    Last edited: Mar 10, 2012
  2. jcsd
  3. Mar 10, 2012 #2


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    You might want to rewrite that. :smile:

    The deal is exactly the surprise - that it looks as if it ought to be convergent and isn't.

    Your integral is the area between the curve and the x-axis, the height of the curve gets smaller and smaller, on the other hand it goes on for ever making more area, so it is not self-evident whether anything like that is convergent or not.

    The usual proof of divergence is that a certain number of successive terms adds up to 1. As you go to the later and later terms, the number of successive terms you have to take to add up to 1 or more gets larger and larger, but you can always get to 1 if you take enough terms, then 1 again, taking even more terms, this for ever so there is no limit to the number of 1's in the sum, so that to total is infinite.
  4. Mar 10, 2012 #3
    oops! I corrected that.

    That's a cool way of explaining divergence, (the bunch of terms add to one, but theres infinite bunches).

    But can't you say the same thing for 1/(x^2)?
    Yet 1/(x^2) is convergent
  5. Mar 11, 2012 #4


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    You can chop up any finite (mathematical) quantity into an infinity of parts, right?

    What do you get when you add together that infinite amount of parts?
    Answer: The finite quantity you started out with..

    Thus, Just because you've got an infinite amount of terms to add, you cannot say merely on THAT information alone whether the sum will be finite or infinite.

    The critical issue to answer that particular question is "how fast" the terms tend towards 0.
  6. Mar 11, 2012 #5
    A set whose sum of the reciprocals diverges is said to be a large set. Similarly, if the sum of the reciprocals converges, the set is said to be a small set. The natural numbers thus make up a large set, while the set of squares is a small set.

    There is a famous conjecture which is still open about small/large sets and there are some counterintuitive (and imo fascinating) examples of small/large sets.

    For example the set of all primes is large, while the set of twin primes is small. (If the twin primes conjecture would be false, this would of course be a trivial conclusion since all finite sets are small.). The Kempner series is quite neat too.

    Check out the wikipedia page about this for more examples.
  7. Mar 11, 2012 #6
    That's a a pretty interesting concept. It makes sense that having an infinite number of terms does not suggests that the sum of these terms is also infinite.

    So in conclusion, we can says the harmonic series is not convergent because it does not approach zero fast enough.

    And our criteria for "fast enough" is "is the integral infinite?"
  8. Mar 11, 2012 #7
    I find this as a convincing argument. Consider:

    S_{2^n} = \sum_{k = 1}^{2^n}{\frac{1}{k}} = \frac{1}{1} + \frac{1}{2} + \left( \frac{1}{3} + \frac{1}{4} \right) + \ldots + \left( \frac{1}{2^{n - 1} + 1} + \ldots + \frac{1}{2^n} \right) \ge \frac{1}{1} + \frac{1}{2} + \frac{2}{4} + \ldots + \frac{2^{n} - 2^{n - 1}}{2^{n}} = 1 + \frac{n}{2}, \ n \ge 1

    Now, the sequence on the r.h.s. is obviously divergent, because it is unbounded. Therefore, by the comparison criterion, the subsequence [itex]\left\lbrace S_{2^n}\right\rbrace[/itex] is divergent as well. Since a subsequence of [itex]\left\lbrace S_n \right\rbrace[/itex] diverges, so does the sequence.
  9. Mar 12, 2012 #8

    I will sketch how [tex]\sum_{n = 1}^{\infty} \bigg(\dfrac{1}{n^2}\bigg) < \ 2[/tex]

    [tex]1 + \dfrac{1}{4} + \dfrac{1}{9} + \dfrac{1}{16} + \dfrac{1}{25} + \dfrac{1}{36} + \dfrac{1}{49} + \dfrac{1}{64} \ + \ ... \ = [/tex]

    [tex]1 + \bigg(\dfrac{1}{4} + \dfrac{1}{9}\bigg) + \bigg(\dfrac{1}{16} + \dfrac{1}{25} + \dfrac{1}{36} + \dfrac{1}{49}\bigg) + \bigg(\dfrac{1}{64} \ + \ ... \ + \dfrac{1}{225}\bigg) \ + \ ... \ = [/tex]

    [tex]1 \ + \ (< \dfrac{2}{4}) \ + \ (< \dfrac{4}{16}) \ + \ (< \dfrac{8}{64}) \ + \ ... \ = [/tex]

    [tex]1 \ + \ (< \dfrac{1}{2}) \ + \ (< \dfrac{1}{4}) \ + \ (< \dfrac{1}{8}) \ + \ ... \ = [/tex]

    [tex]1 \ + \ (< 1) \ = [/tex]

    [tex](< 2)[/tex]
    Last edited: Mar 12, 2012
  10. Mar 12, 2012 #9
    i was not aware that you could do that (1/4 + 1/9 = <1/2).
    that was a nice a proof, but i can easily show that 1/x^2 is convergent by showing its integral is convergent.

    what i meant by post is that

    1/x^2 tends to zero
    1/x tends to zero

    1/x^2 has a graph that looks like it is convergent
    1/x has a graph that looks convergent

    someone posted suggesting that even though as 1/x approaches infinity, it becomes very small but still adds to infinity.

    1/x^2 gets very small, but you could add a million small fractions together and get one again.

    I don't quite follow what the second series is. I understand that you are creating a smaller sequence, and showing that the smaller sequence is divergent which suggests that the larger sequence (1/x) must also be divergent.

    i don't see how you created that "subsequence"
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