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.

  1. Show that $k < m_1 + 2$.
  2. Give an example of such a sequence of length $k$ for any positive integer $k$.

Any kind of help would be appreciated. 😊


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.


Let the numbers $1/m_i$ be $1/N,2/N,3/N,...,k/N$ for suitable $N$.

  • $\begingroup$ Would it be a proper construction if we cannot specify any methods to find N? $\endgroup$ – MathIsNice1729 Oct 24 '15 at 16:23
  • $\begingroup$ I suggest $N=k!$ $\endgroup$ – Empy2 Oct 24 '15 at 16:50
  • $\begingroup$ Hmm... This is what I needed... $\endgroup$ – MathIsNice1729 Oct 26 '15 at 6:35

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