QOTD (from an old IMO): show that none of these sums are integers

Here’s an old IMO question (or maybe it’s off one of the short or long lists) that I haven’t made any progress on:

Show that for any natural numbers \(a\) and \(n\) the sum
\[
1 + \frac{1}{1+a} + \frac{1}{1 + 2a} + \cdots + \frac{1}{1+n a}
\]
is not an integer.

An obvious and cool corollary is that the harmonic numbers are never integers.

  • I can give this a try, if you want still.

    This week, I have a bit of spare time.

  • If you get time, there is an interesting probability problem (a) on page 7 (here: http://isites.harvard.edu/fs/docs/icb.topic1025991.files/8%20-%20ConditionalProbability.pdf) about “true” prophets.

    See what you make of it.

  • swiftset

    Sure, give it a shot. I haven’t worked on it in a while, but I think a fruitful approach might be to try and sandwich it with log-like integrals and then somehow (this is the bit I haven’t been able to do) argue that there aren’t any integers between the two integrals.

    The true prophets problems look like applications of Bayes’ theorem.

  • Whoa!

    This problem, or the sum described, turns out to be a much studied.

    My original proof
    http://tinyurl.com/czleqr4

  • swiftset

    The proof doesn’t work as is: I agree with your first observation about no two terms adding up to larger than 1 unless one term is 1, but you can definitely add up more terms, none of which are 1, and get something larger than 1. You know this because the harmonic number \(H_n\) is on the order of \(\log n\), so the sequence diverges.

    I didn’t know that there was a standard proof that the harmonic numbers aren’t integers, but maybe we can adapt that for general \(a\)? It’s worth a try.

    • Sorry for the lat response. Distracted by schoolwork.
      Anyway, I see the serious error in my proof.

      I thought about it and came up with another explanation. But when I go to edit my proof, my document is nowhere to be found. As if been deleted: everything including drafts, final copy that was both on my computer and hosted on website.
      Curious.

      Anyway, my quick work-around is this.

      The crux of the problem is wanting to show that any partial sums in the finite series cannot add to an integer.

      That means the numerator, which is a sum of the integers from the denominator from the original terms, and divided by the denominator, which is the product of said integers, must equal some integer value.

      This happens iff with perfect numbers, only case are the first three numbers in series: 1,2, & 3. Never works thereafter.

      Applying the Fundamental Theorem of Arithmetic shows that it is impossible to pull out a common factor from the numerator that can cancel out with a possible common factor from the denominator.

      I hope this is clear. I am too lazy to re-type what I original wrote from scratch, again.