Proof related to Harmonic Progression

Question

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.

Answer 1

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 $$ \frac{1}{m_1}-(k-1)\frac{1}{m_1(m_1+1)} = \frac{1}{m_k} >0 \\ \frac{1}{m_1} > (k-1)\frac{1}{m_1(m_1+1)} \\ 1 > (k-1)\frac{1}{m_1+1} \\ m_1+1 > k-1 \implies k < m+2 $$

Answer 2

Since they are in arithmetic progression let us call $1/m_n-1/m_{n+1}=L>0$. We then have that $$\begin{align} \sum_{n=1}^{k-1}\left(\frac{1}{m_n}-\frac{1}{m_{n+1}}\right) &= \sum_{n=1}^{k-1}L \\ &= (k-1)L \\ &= \frac{1}{m_1}-\frac{1}{m_k} \end{align}$$ Which means we just have to show $$k=\frac{\frac{1}{m_1}-\frac{1}{m_k}}{L}+1<m_1+2\qquad (1)$$ Multiply both sides by $m_1$ $$\frac{1-\frac{m_1}{m_k}}{L}<m_1(m_1+1)$$ Multiply by $L=1/m_1-1/m_2$ and move around some terms $$\begin{align} 1 &<Lm_1(m_1+1)+\frac{m_1}{m_k} \\ &= \left(\frac{1}{m_1}-\frac{1}{m_2}\right)m_1(m_1+1)+\frac{m_1}{m_k} \\ &=m_1+1-\frac{m_1(m_1+1)}{m_2}+\frac{m_1}{m_k} \end{align}$$ Hence we for $(1)$ to be true we only need to show $$\begin{align} 0 &< m_1+\frac{m_1}{m_k}-\frac{m_1(m_1+1)}{m_2} & (2) \\ &= m_1+\frac{m_1}{m_k}-\frac{m_1(m_1+1)}{m_1+x} & x\ge 1 \end{align}$$ Notice that because $x\ge 1$ we have $$\begin{align} m_1+\frac{m_1}{m_k}-\frac{m_1(m_1+1)}{m_1+x} &\ge m_1+\frac{m_1}{m_k}-\frac{m_1(m_1+1)}{m_1+1} \\ &=\frac{m_1}{m_k} \\ &> 0 \end{align}$$ Thus we see that $(2)$ holds and tracing back our work we see that this implies our original inequality, $(1)$, holds.

Answer 3

It is important in this problem to understand that the maximun number $k$ is determined by the choice of the two first integers $m_1,m_2$.

Let $m_1$ and $m_2=m_1+h$ (where $h\ge 1$); the common difference of the a. p. is $d=\frac{-h}{m_1(m_1+h)}$ so we have $$u_1=\frac{1}{m_1}\\u_2=\frac{1}{m_1+h}\\u_3=\frac{1}{m_1+h}+\frac{-h}{m_1(m_1+h)}=\frac{m_1-h}{m_1(m_1+h)}\\u_4=\frac{m_1-2h}{m_1(m_1+h)}\\u_5=\frac{m_1-3h}{m_1(m_1+h)}\\....\\....\\u_k=\frac{m_1-(k-2)h}{m_1(m_1+h)}$$ The progression stop when $u_k$ is not anymore positive. Meanwhile one has $$m_1-(k-2)h\gt 0\iff m_1\gt(k-2)h\ge k-2$$ Thus $$k\lt m_1+2$$