Q:

Proof related to Harmonic Progression

The question is as follows:

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. Then prove that $k<m_1+2$.

I am not sure how to tackle this problem. I tried taking the common differences, but I can't seem to be able to arrive at the final expression. Any help is appreciated.

A:

Assume the arithmetic progression, starting with $\frac1{m_1}$, contains $k$ positive terms.  Now the (uniform) term difference is 
$$
d = \frac{1}{m_1} - \frac{1}{m_2}
\\ m_2 \geq m_1+1 \implies d \geq \frac{1}{m_1} - \frac{1}{m_1+1}=\frac{1}{m_1(m_1+1)}
$$
The entire sequence, which ends in a positive value $\frac{1}{m_k}$, consists of $k-1$ steps of size $d$.
Thus
$$
\frac{1}{m_1}-(k-1)\frac{1}{m_1(m_1+1)} = \frac{1}{m_k} >0 \\
\frac{1}{m_1} > (k-1)\frac{1}{m_1(m_1+1)}
\\ 1 > (k-1)\frac{1}{m_1+1}
\\ m_1+1 > k-1 \implies k < m+2
$$

