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Let $m_1<m_2<m_3<\cdots<m_k$ be postive integers such that $\frac{1}{m_1}$, $\frac{1}{m_2}$, $\frac{1}{m_3}$, $\cdots$, $\frac{1}{m_k}$ are in arithmetic progression.

I've seen plenty of proofs to show that $k < m_{1} + 2$ (see here), and that makes sense to me. I was wondering if there is a more simple proof for a less strenuous claim however - simply that there doesn't exist any such infinite sequence. What would be the most direct way of asserting this without going the extra step and showing there are at most $m_{1} + 1$ integers, should one exist?

I've tried employing induction, or just showing that the representation in terms of $m_{1}$ and $m_{2}$ cannot lead to a valid $m_{n}$, but have been having a harder time making progress than just directly showing $k < m_{1} + 2$.

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Note that we have $0 < \frac{1}{m_i} < 1$. However for them to be in an arithmetic sequence we must have the difference between each one is $1 > \frac{1}{m_1} - \frac{1}{m_2} > 0$. Call this quantity $\delta$.

Then set $N = \lceil \delta^{-1} \rceil$. We have the $k$th element of your sequence is $\frac{1}{m_1} - (k-1)\delta$. So the $N$th element is $\frac{1}{m_1} - (N-1)\delta < \frac{1}{m_1} - 1 < 0$. This contradicts the form of the elements in the sequence (namely that they are all positive).

The point is given any decreasing arithmetic sequence, if it is infinite it must eventually go below 0.

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