Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I've been looking at unit fractions, and found a paper by Erdos "Some Properties Of Partial Sums Of The Harmonic Series" that proves a few things, and gives a reference for the following theorem:

$$\sum_{k=0}^n \frac{1}{m+kd}$$ is not an integer.

The source is:

Cf. T. Nagell, Eine Eigenschaft gewissen Summen, Skrifter Oslo, no. 13 (1923) pp. 10-15.


Question

Although I would like to find this source, I've checked with my university library and it seems pretty out of reach. What I'm really hoping for is a source that's more recent or even written in English.

Finding this specific source isn't everything, I'll be fine with pointers to places with similar results.

share|improve this question
2  
Assuming m, n, and d are natural numbers, do you just want to use this result? I think it would not be too hard to prove. I can't help you find a reference, though. Did you ask about interlibrary loan? Librarians have helped me find such things in that past, including a German paper that I'd almost given up hope of seeing. –  Eric Tressler Sep 19 '10 at 20:24
    
I did check with them, they told my professor that they would do some looking for us. Since they did not sound hopeful, I thought I would check here. –  mmm Sep 20 '10 at 1:43
add comment

3 Answers

up vote 2 down vote accepted

You can cite H Belbachir and A Khelladi, On a sum involving powers of reciprocals of an arithmetic progression, Ann Math Inform 34 (2007) 29-31, MR 2009d:11018, where a more general result is given. If you are OK with Russian, there is Z D Gorskaya, On an arithmetic property of a harmonic sum, Ukrain Mat Z 6 (1954) 375-384, MR 16, 998j.

Nagell wrote a nice intro Number Theory textbook, which was republished by the American Math Society. Maybe the result is in it.

EDIT: I have had a look at the Belbachir and Khelladi paper, at http://www.kurims.kyoto-u.ac.jp/EMIS/journals/AMI/2007/ami2007-belbachir.pdf and I find that it rests heavily on the Shorey and Tijdeman paper cited in Gjergji Zaimi's answer.

FURTHER EDIT: I think that Erdos himself proves the result in a paper of 1932 (but it's in Hungarian), Egy Kurschak-fele elemi szamelmeleti tetel altalanositasa, Lapok 39 (1932) 17-24 (my apologies for omitting the numerous diacritical marks). This is freely available, with a summary in German at the end, at http://www.renyi.hu/~p_erdos/1932-02.pdf It would seem that in 1932 Erdos was unaware of Nagell's work.

share|improve this answer
add comment

I tracked down Nagell's paper in early 1992, since I had found a proof and wanted to see whether it was the same as his. (It turned out to be essentially the same idea.) Unfortunately, I've since lost my photocopy of his paper, but it came from UIUC, so that would be where I'd start looking. If I remember right, the journal where it appeared is incredibly obscure, and UIUC was the only place in North America that had a copy.

Here's a proof copied from my very old TeX file. A bit awkwardly written, but it explains how to do the case analysis, which is the part that makes this approach simpler than what Erdős did.

Suppose that $a$, $b$, and $n$ are positive integers and $$\frac{1}{a}+\frac{1}{a+b}+\frac{1}{a+2b}+\cdots+\frac{1}{a+nb} = c \in \hbox{\bf Z}$$ We can take $\gcd(a,b) = 1$ and $a > 1$, and it is easy to show that $n > 2$.

Suppose that $b$ is odd. In the arithmetic progression, there is then a unique number $a + mb$ divisible by the highest possible power of $2$, because for all $k$, the progression runs through a cycle modulo $2^k$ which contains each value exactly once. Then multiplication by $\ell = \hbox{lcm}(a,a+b,\ldots,a+nb)$ gives $$\frac{\ell}{a}+\frac{\ell}{a+b}+\frac{\ell}{a+2b}+\cdots+\frac{\ell}{a+nb} = \ell c.$$ All of the terms here except $\frac{\ell}{a+mb}$ are even. Thus, $b$ cannot be odd.

Now suppose that $b$ is even, and $b \le \frac{n-2}{3}$. By Bertrand's Postulate, there is a prime $p$ such that $\frac{n+1}{2} < p < n+1$. Then $p$ does not divide $b$, and $p$ is odd. We must have $a \le n$, because $$1 \le c = \frac{1}{a}+\frac{1}{a+b}+\cdots+\frac{1}{a+nb} < \frac{n+1}{a}.$$

Since $b$ generates the additive group modulo $p$ and $n+1 > p$, at least one of the numbers $a$, $a+b$, \ldots, $a+nb$ is divisible by $p$. At most two are, since $2p > n+1$. Suppose that $p$ divides only the term $a+kb$. Then $$\frac{1}{a+kb} = c-\sum_{j \neq k}{\frac{1}{a+jb}}.$$ The denominator of the left side is divisible by $p$, but that is not true of the right side. Thus, $p$ must divide two terms.

Now suppose that $p$ divides $a+{\ell}b$ and $a+kb$, with $mp = a+{\ell}b < a+kb = (m+b)p$. Then $$\frac{1}{p}\left(\frac{2m+b}{m(m+b)}\right) = c - \sum_{j \neq \ell,k}{\frac{1}{a+jb}}.$$ This implies that $p \mid (2m+b)$. However, $$a+{\ell}b \le n + (n-p)\left(\frac{n-2}{3}\right) < n + \frac{n}{2}\left(\frac{n-2}{3}\right).$$ Therefore, $$m < \frac{a+{\ell}b}{n/2} < \frac{n-2}{3} + 2.$$ It follows that $2m+b \le n+1$. However, $n+1 < 2p$, so $2m+b = p$. This contradicts the fact that $b$ is even and $p$ is odd.

Finally, suppose that $b$ is even, and $b \ge \frac{n-1}{3}$. Since $a > 1$ and $a$ is odd, we must have $a \ge 3$. We must also have $b \ge 4$, since if $b=2$, then Bertrand's Postulate and the fact that $n \ge a$ (as above) imply that one of the terms is a prime, which does not divide any other term.

Now, we show that $n \le 13$. To do that, note that $$\frac{1}{a}+\frac{1}{a+b}+\cdots+\frac{1}{a+nb} < \frac{1}{3} + \frac{1}{7} + \frac{3}{n-1}\left(\frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}\right).$$ A simple computation shows that for $n \ge 14$, $$\frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n} < \frac{11}{63}(n-1),$$ which implies that the sum is less than 1. Hence, we must have $n \le 13$.

Thus, the sum is at most $$\frac{1}{3} + \frac{1}{7} + \frac{1}{11} + \cdots + \frac{1}{47} + \frac{1}{51} + \frac{1}{55} < 1.$$

Therefore, the sum is never an integer, and the theorem holds.

share|improve this answer
    
+1. This was surprisingly elementary and quick! :) –  Gjergji Zaimi Dec 19 '13 at 3:28
add comment

If you are willing to use some heavier results, the fact that $$\sum_{k=0}^{n}\frac{1}{m+kd}$$ is never an integer follows from the following theorem by Shorey and Tijdeman (which refines a theorem of Sylvester):

The greatest prime factor of the product $m(m+d)\cdots (m+nd)$ is greater than $n+1$ unless $(m,d,n)=(2,7,2).$

This is proven in "On the greatest prime factor of an arithmetical progression", and refinements are given in subsequent papers of the authors, finding these references shouldn't be that hard. It implies your result because it shows that one of the fractions has a denominator divisible by a prime $p$ which doesn't appear in any other denominator.

share|improve this answer
    
I don't have a copy of the latter paper, but I wonder whether the result generalized is from Sylvester's "On arithmetical series," which has a theorem that easily implies Bertrand's Postulate (cf. mathoverflow.net/questions/103736/…) which was used in Henry Cohn's proof above. –  Benjamin Dickman Dec 22 '13 at 21:14
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.