3

In the post “Is there an elementary proof that ∑k=1/k is never an integer?” there is a simple and elegant proof by Bill Dubuque, who uses the prime 2 as a basis for his proof. I wondered whether a similar proof (perhaps less simple, but still using simple steps) can be given using any other, i.e. odd, prime.

I start with a prime p and consider the sum $1+1/2+1/3+ + 1/p=H(p)$ and multiply both sides with p! to get $p!/1+p!/2+..+p!/p=H(p)p!$ and then I consider the quantities (mod p)

All the terms p!/k vanish except p!/p so that p!/p=(p-1)!=H(p)p!. Assuming H(p) is an integer then we have (p-1)!=0 which a is a contradiction since (p-1)! cannot be 0 (according to Wilsons theorem it is actually equal to -1 (mod p)).

The next step is to consider H(pq) where p and q are primes and $q<p$. We consider the sum $1+1/2+1/3+ + 1/(pq)=H(pq)$. We multiply with $(pq)!/p^{q-1}$ to get

$(pq)!(p^{q-1})(1/1+1/2+...+1/(pq))=H(pq)(pq)!/p^{q-1} $

Again most of the LHS terms vanish except those with a factor p in the denominator, leaving the terms

$(pq)!/p^{q-1}(1/p+1/(2p)+…+1/(pq))=((pq))!/p^{q-1})(1+1/2…+1/q)=((pq)!/p)H(q)$ to arrive at the interesting “multiplication formula”:

$((pq)!/p^{q-1})H(q)=H(pq)(pq)!/p^{q-1}$ (mod p)

Edit(2): In order to have integers on both sides on the equations above I need to multiply both sides with (pq)! Edit 3: I divide $(pq)!$ with $/p^{q-1}$ rather than p since (pq)! contains q factors of p, that is the factor $p^q$. This will make all terms vanish (mod p) except 1/p, 1/(2p),...and /1(qp).

If H(pq) is an integer then the LHS must be an integer in contradiction to the proof above for a prime.

I realize I still need to cover the case p=q and products of multiple primes, but my question is whether I am OK so far, in particular with my formula for H(pq)? The internet have some treatments of multiple harmonic sums often mentioning the complicated term “p-adic” , unfortunately not really readable for amateurs.

Edit 3:I believe my comment below to Robinson gives a proof but not the systematic investigation alternative I started on.

Community
  • 1
Mikael Jensen
  • 1,817
  • 3
    $p$-adics are terrific for amateurs. See my paper with Alf van der Poorten in the American Math Monthly from about 20 years ago. – Gerry Myerson Mar 17 '15 at 12:14
  • 1
    It is a little ironic that your the proof of your result in the prime case is easily generalized to a proof of the whole result using Bertrand's postulate, ( choose a prime $p$ with $\frac{n}{2} < p <n,$ and see what happens when you multiply through by $n!$, which Bill Dubuque's proof was an answer to, in response to the question whether there was a simpler proof than the BP one. – Geoff Robinson Mar 19 '15 at 10:15
  • @Robinson With Bertrands theorem it becomes easy when I know there is no additional multiple of p in the interval, but I actually don’t think I need it. I can also take an arbitrary n and prime p and consider a=[n/p]. n! will contain the factor p^a as the term with the highest exponent of p. The sequence 1/1…1/n, must contain a highest exponent b of p, so I multiply both sides with $ n!/(p^{a-b}) $ which will leave $ (n!/(p^{a-b}))/p^b$ as the only term which doesn’t contain p and therefore different from 0 (mod p). – Mikael Jensen Mar 20 '15 at 09:00
  • IT occurs to me that we must multiply with the Least Common denominator LCD and not the (pq)! etc I have used. (p)! have too many factors of p, making for instance (1+1/2+.. +1/p)p! =0 (mod p). It works for (1+1/2+.. +1/p)LCD which is not 0 (mod p) and H(p)*LCD =0 if H(p) is an integer. – Mikael Jensen May 23 '15 at 14:15

0 Answers0