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. 


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*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. 

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.
