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.

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

  • $\begingroup$ 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 $3 < 6 + 2$. $\endgroup$ – 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 $x>0$.)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.