Sequence of positive integers such that their reciprocals are in arithmetic progression Let $m_1 < m_2 < \ldots < m_k$ be $k$ distinct positive integers such that their reciprocals $\dfrac{1}{m_i}$ are in arithmetic progression. 


*

*Show that $k < m_1 + 2$.

*Give an example of such a sequence of length $k$ for any positive integer $k$.


Any kind of help would be appreciated. 
 A: I recently stumbled across a very similar problem and wanted to share my thoughts.
The problem with using $k!$ means that the maximal progression of the set may not be $k$ (the set can sometimes be extended past $k$ terms), which is what my question involved. For example, the set $\left(\frac{1}{5!},\frac{2}{5!},\frac{3}{5!},\frac{4}{5!},\frac{5}{5!}\right)$ can be extended to a sixth term of $\frac{6}{5!}$. The only times we can use $k!$ as the denominator for a maximal progression of $k$ is when $k$ is one less than a prime or less than or equal to $3$.
The question concerned whether it is possible to construct sets of this kind where the maximal progression can be any positive integer $n$. In fact, there are infinitely many maximal progressions for each positive integer $n$:
Due to Dirichlet's Theorem, there are infinitely many primes that are $1\pmod{n}$, so let $p=1+mn$ where $p$ is prime.
Let $$k=\prod_{r=0}^{n-1}{(1+mr)}=1(1+m)(1+2m)\ldots(1+m(n-1))$$
All of $\frac{1}{k}$, $\frac{1+m}{k}$, $\frac{1+2m}{k}$, $\ldots$, $\frac{1+m(n-1)}{k}$ are reciprocals of positive integers, as the numerator is a factor of the denominator and the difference between each term is $\frac{m}{k}$.
$\frac{1}{k}-\frac{m}{k}\le 0$ and so is not the reciprocal of a positive integer as it is negative, and $\frac{1+m(n-1)}{k}+\frac{m}{k}=\frac{1+mn}{k}=\frac{p}{k}$, and since $p$ is prime and so is relatively prime with each factor of $k$, $\frac{p}{k}$ is not the reciprocal of a positive integer either and so this is a maximal progression of $n$.
This means we have generated one set of reciprocals of positive integers in arithmetic progression for one unique value of $p$ for each $n$. But since there are infinitely many $p$ that work, there are infinitely many solutions for which this works.
A: Let the numbers $1/m_i$ be $1/N,2/N,3/N,...,k/N$ for suitable $N$.
A: Hint for the bound: Use the fact that the common difference $d$ of the AP is $\ge \frac{1}{m_1}-\frac{1}{m_1+1}$. Now if we take $(n-1)$ steps of length  down from $a=\frac{1}{m_1}$, the result must be $\gt 0$.
Hint for the construction: The first interesting length is $k=3$. There we have the familiar AP $\frac{1}{2},\frac{1}{3},\frac{1}{6}$. Now the only challenge is constructing an AP of length $4$.
We look for an AP of the shape $\frac{1}{m},\frac{1}{2q},\frac{1}{3q},\frac{1}{6q}$. 
So we want $\frac{1}{m}-\frac{1}{2q}=\frac{1}{6q}$, that is, $\frac{1}{m}=\frac{4}{6q}$. Take $q=4$.So our first term is $\frac{1}{6}$, and the others are $\frac{1}{8}$, $\frac{1}{12}$, and $\frac{1}{24}$. 
Now on to $5$. In our example of length $4$, we had $q=4$, so common difference $\frac{1}{24}$ and  first term  $\frac{1}{6}$. Multiply the denominators of the length $4$ example by a new suitable $q$. We want to insert a new first term $\frac{1}{m}$, where $\frac{1}{m}-\frac{1}{6q}=\frac{1}{24q}$. Note that $q=5$ works. The new common difference is $\frac{1}{120}$ and the new first term is $\frac{1}{24}$.
Continue. At each stage we multiply the denominators of the length $l$ AP by a suitable $q$, and insert a new first term. Same calculation.
Now we can look back on the examples, and write down a simple explicit formula. 
