4
$\begingroup$

Question: If $A\geq B>0$ are rational and $x_{1}\leq x_{2}\leq \cdots \leq x_{n}$ are integers such that $A\geq \sum_{j=1}^{n}\frac{1}{x_{j}}\geq B$, then what is an upper bound on $x_{j}$ in terms of $x_{1},x_{2},\ldots, x_{j-1},A,B,$ and $n$? Is the bound sharp?

If $A=B$, then such a bound is provided in E. Landau Uber die Klassenzahl der binaren quadratischen Formen von negativer Discriminante, Math. Ann. 50 (1903), 671--676; doi: 10.1007/BF01444311, eudml. Specifically one has:

Theorem: If $x_{1}\leq x_{2}\leq\cdots\leq x_{n}$ are integers such that $\sum_{j=1}^{n}\frac{1}{x_{j}}=R$, then $x_{j}\leq (n-i+1)/R_{j-1}$ where $R_{0}=R$, and $R_{k}=R_{k-1}-\frac{1}{x_{k}}$.

$\endgroup$

1 Answer 1

1
$\begingroup$

UPDATE. As pointed out in the comments, an upper bound exists for $x_i$ only if $\sum_{j=1}^{i-1} \frac{1}{x_j} < B$. If we have $B\leq \sum_{j=1}^{i-1} \frac{1}{x_j}<A$, then any sufficiently large $x_i,x_{i+1},\dots,x_n$ will work fine.


First, notice that the theorem is rather trivial. Indeed, for any $i=1,2,\dots,n$, we have: $$R = \sum_{j=1}^n \frac{1}{x_j} \leq \sum_{j=1}^{i-1} \frac{1}{x_j} + \sum_{j=i}^{n} \frac{1}{x_i} = \sum_{j=1}^{i-1} \frac{1}{x_j} + (n-i+1)\frac{1}{x_i},$$ implying that $$x_i \leq \frac{n-i+1}{R-\sum_{j=1}^{i-1} \frac{1}{x_j}}.$$

Replacing "$R = \sum_{j=1}^n \frac{1}{x_j}$" with "$B \leq \sum_{j=1}^n \frac{1}{x_j}$", we get a similar upper bound for $x_i$ (assuming that $\sum_{j=1}^{i-1} \frac{1}{x_j} < B$): $$x_i \leq \frac{n-i+1}{B-\sum_{j=1}^{i-1} \frac{1}{x_j}}.$$

The bound is tight: it is attained when $\sum_{j=1}^n \frac{1}{x_j} = B$ and $x_i=x_{i+1}=\dots=x_n$.

The value of $A$ can be used to obtain a lower bound for $x_i$. Namely, from $$A\geq \sum_{j=1}^n \frac{1}{x_j} \geq \sum_{j=1}^i \frac{1}{x_j}$$ we get $$x_i \geq \frac{1}{A - \sum_{j=1}^{i-1} \frac{1}{x_j}}.$$

P.S. There is a related Kellogg problem, which for the case $R=1$ bounds $x_n$ uniformly as $x_n \leq s_{n-1}-1$ (where $s_j$ is Sylvester's sequence). This bound can be also extended to lower terms, giving $x_i \leq (n+1-i)\cdot (s_{i-1}-1)$ for all $i=1,2,\dots,n$.

$\endgroup$
2
  • 1
    $\begingroup$ There doesn't seem to be any guarantee that $B-\sum_{j=1}^{i-1}1/x_{j}>0$. In particular, you may be dividing by a negative number, or worse 0. Perhaps I am missing your point? $\endgroup$
    – Paul
    Jan 16, 2017 at 23:20
  • 1
    $\begingroup$ @Paul: That's a valid concern. The thing is though that if $B-\sum_{j=1}^{i-1}\frac{1}{x_j}\leq 0$, then there is no upper bound for $x_i$ exists -- it can take any sufficiently large values. I've added this comment to my answer. $\endgroup$ Jan 17, 2017 at 2:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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