Let $\{r_i\}_{i \in \mathbb{N}}$ be a periodic sequence of integers. That is, there exists a $t$, such that, for all $i \in \mathbb{N}$, $r_i = r_{i+t}$. What I need is a large prime dividing the numerator of $\displaystyle \sum_{i=1}^n \dfrac{r_i}{i}$. I general, I'd like to know more about the largest prime dividing that sum as $n$ goes to infinity. In particular, I would love to know a proof (or even more spectacular, a counterexample) to the following conjecture:

For every $c \in \mathbb{R}$, there exists a positive integer $n \ge 2$ and a prime $p > cn$ that divides the numerator of $\displaystyle \sum_{i=1}^n \dfrac{r_i}{i}$

Furthermore, I metaconjecture that my effort over the last 6 months to prove this, will be nullified within an hour or two.

If at least one of my conjectures turns out to be false, for which $c > 0$ is my first conjecture true and easily provable?

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    $\begingroup$ It would appear that your metaconjecture has been refuted. $\endgroup$ Apr 5, 2011 at 22:55
  • $\begingroup$ Do you have any evidence/work that makes you think this is interesting? Especially wrt to obtaining some useful bounds on the numerator. Your conjecture appears prima facie 'hard' especially since there are no easy counterexamples despite the fact that there is no bound on your $c \in \mathbb{R}$ or on the choice of periodic sequence. I'd be interested to see what led you to think it is true. $\endgroup$
    – Chuck
    Apr 5, 2011 at 23:27
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    $\begingroup$ I believe that it is well-known (perhaps in Gouvea's "p-adic Analysis") that with $r_i=1$, each prime divides the numerator for only finitely many $n$. $\endgroup$ Nov 4, 2011 at 20:47
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    $\begingroup$ @Kevin indeed; arxiv.org/PS_cache/arxiv/pdf/1102/1102.0765v1.pdf $\endgroup$
    – Woett
    Nov 4, 2011 at 23:57

1 Answer 1


This is partially an answer to Chuck.

The denominator is divisible by all primes $p$ such that $\dfrac{n}{\log{n}} \ll p < n$, so the denominator grows exponentially in $n$. Furthermore, I believe (and have been too lazy and incompetent to check) that the sum is (in absolute value) $\gg n^{-t}$. This implies that the numerator also grows exponentially in $n$. So to me this shows that the problem should be easy; how can there ever not be a prime $> cn$ dividing somethig like $e^n$? This is espeically true for $c=1$, since if we know that the numerator is only divisible by primes smaller than $n$, we actually know that all the prime divisors of it must be $\ll \dfrac{n}{\log{n}}$. So, since Gerry pointed out that my metaconjecture has been refuted, I'd like to coin a new one: for $c=1$ the conjecture is provable by a simple counting argument.

  • $\begingroup$ The product of all the primes up to $n/\log n$ is only $\exp(O(n/\log n))$, true, so the only way the $c=1$ conjecture could fail is if these smallish primes occurred to higher and higher powers (on average). $\endgroup$ Nov 4, 2011 at 19:57
  • $\begingroup$ Indeed, I should have mentioned that explicitly. $\endgroup$
    – Woett
    Nov 5, 2011 at 1:54

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