# More simple proof of reciprocal sequence

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 $k < m_{1} + 2$.

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).