# 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.

• What is AP? Arithmetic progression? – Arthur May 12 '15 at 16:44
• Yes, AP-Arithmetic Progression. – Swadhin May 12 '15 at 16:45
• Note that you need these integers to be distinct, though this is just a little technical point. – Wojowu May 12 '15 at 16:54
• To find "such a sequence", take $k=1$ and let $m_1$ be any positive integer. – Théophile May 26 '15 at 3:24

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.

• One example sequence would be $6, 3, 2$. That way, the reciprocals form the sequence $\frac{1}{6}, \frac{1}{3}, \frac{1}{2}$, and $3 < 6 + 2$. – 2012ssohn May 29 '15 at 3:22

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.

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 $x>0$.)