30
$\begingroup$

I'm primarily a programmer, so forgive me if I don't know the proper nomenclature or notation.

Last night, an old teacher of mine told me about a question that had caused some noodle-scratching for him:

For any two sequences of consecutive integers, can the sums of their reciprocals be equal?

Now, I gather that these sums are called "harmonic numbers" if the consecutive sequence begins with 1. But what if it doesn't begin with 1?

How might we go about proving that for any two such sequences, their sums are unequal?

I have written a quick Python script that returns, in the form of a reduced fraction, the sum of any $\sigma(m, k)$ where $m$ is the first number to consider and $k$ is the length of the sequence.

So, I guess I have two questions:

1) Where can I find out about the current state of research on this question?

2) What are the most likely approaches, or "hooks" that I might grasp onto to arrive at a proof?

(Via the comments, I'll add the following for clarity:)

$\sigma(m,k)$ is $\frac{1}{m} + \frac{1}{m+1}+\dots +\frac{1}{m+k−1}.$ My question is: Can there exist distinct pairs $(m_1,k_1)$ and $(m_2,k_2)$ such that the corresponding sums are equal?

  • 0
    Mariano and I have evidently interpreted the question in two different ways, and it's not clear that either of us has it right. Could you be a little more specific? Do you, for instance, intend (i) finite sums of (ii) reciprocals of *consecutive* integers?2011-02-14
  • 0
    Yes yes, consecutive integers - I'll edit my post. I'm not sure what you mean by 'finite sums' in this case.2011-02-14
  • 0
    just that the sum stops after a certain point (as it must if you include consecutive integers: the infinite harmonic series diverges).2011-02-14
  • 4
    Your sigma(m,k) is $\frac1m + \frac1{m+1} + \dots \frac1{m+k-1}$, is that right? (This sum can also be written as $H_{m+k-1} - H_{m-1}$.) And your question is whether there can exist distinct pairs $(m_1, k_1)$ and $(m_2, k_2)$ such that the corresponding sums are equal, i.e., $H_{m_1+k_1-1} - H_{m_1-1} = H_{m_2+k_2-1} - H_{m_2-1}$. Is that right?2011-02-14
  • 1
    ShreevatsaR: That's exactly right.2011-02-14
  • 0
    Recently posted to MO, http://mathoverflow.net/questions/260933/are-all-partial-consecutive-harmonic-subsums-distinct2017-02-03
  • 0
    It appears that the result is due to Erdos and Niven, Some properties of partial sums of the harmonic series, Bull Amer Math Soc 52, No. 4 (1946) 248-251, available by way of http://www.ams.org/journals/bull/1946-52-04/home.html2017-02-08

3 Answers 3