Integer reciprocals in arithmetic progression 
Let $m_1,m_2,\ldots,m_k$ be $k$ positive integers such that their reciprocals are in AP. Show that $k<m_1+2$. Also find such a sequence.  

Whatever way I tried, whichever formula I used, I could not eliminate the $m_k$ term, which I reckon would be the ideal scenario to find the solution. I am exhausted of any further ideas, please help. Thank you.
 A: Note that we can, w.l.o.g., assume that $m_1<m_2<...<m_k$ (even though we specifically have $m_1$ in the bound, check yourself why we can assume that). Now we have $m_2\geq m_1+1$, so that $\alpha=\frac{1}{m_1}-\frac{1}{m_2}\geq \frac{1}{m_1}-\frac{1}{m_1+1}=\frac{1}{m_1(m_1+1)}$. Now, as $\alpha$ is a difference between some two consecutive reciprocals, it must be the difference between any two consecutive reciprocals (as we assume they form AP). This means that we will have $\frac{1}{m_i}=\frac{1}{m_1}-(i-1)\alpha\leq \frac{1}{m_1}-\frac{i-1}{m_1(m_1+1)}=\frac{m_1+1-i+1}{m_1(m_1+1)}=\frac{m_1+2-i}{m_1(m_1+1)}$. But every term of this sequence must be positive, so the bound we have just found has to be positive as well, which means that $m_1+2-i>0, i<m_1+2$. In particular, this bound has to hold for $i=k$.
I'm not entirely sure what is meant with "find such a sequence", so I am going to leave it to you, unless you are able to clarify that.
A: Assume that the $\{m_i\}$ are in ascending sequence, so that $m_1$ is the minimum value. Then consider the increment in the arithmetic progression (AP) of the reciprocals by looking at the second integer term, $m_2$. The smallest possible value for this is $m_1+1$. This gives an increment in the AP, $d$, of 
$$d = \frac{1}{m_1+1}-\frac{1}{m_1} = \frac{-1}{m_1(m_1+1)}$$
Using this increment - the smallest possible - would give an $(m_1+2)$th term in the AP of: 
$$\frac{1}{m_1} + (m_1+1)d = \frac{1}{m_1} +\frac{-(m_1+1)}{m_1(m_1+1)} = 0$$
and zero is of course not the reciprocal of an integer, so there can be at most  $(m_1+1)$ terms.
This gives $k<m_1+2$ as required.
A: An improvement (still far from the optimal estimate.)
By the convexity of the function $1/x$ we can see that $m_2-m_1<m_3-m_2<\ldots<m_k-m_{k-1}$, which provides $m_{i+1}-m_i\ge i$ and therefore $m_i\ge m_1+\frac{i(i-1)}2$.
Let $2\le\ell\le k$. Then
$$
\frac{\frac1{m_1}-\frac1{m_\ell}}{\ell-1}
= \frac{\frac1{m_1}-\frac1{m_k}}{k-1}
< \frac1{(k-1)m_1} 
$$
$$
(k-1)m_1 > (k-\ell)m_\ell \ge (k-\ell)\left(m_1+\frac{\ell(\ell-1)}2\right)
$$
$$
m_1 > \frac{(k-\ell)\ell}2
$$
so
$$
m_1 \ge \frac{(k-\ell)\ell+1}2.
$$
Chosing $\ell=\lceil k/2\rceil$ we can get
$$
k<\sqrt{8m_1}.
$$
(By the AM-GM, $\sqrt{8x}\le x+2$ for $x>0$.)
