We have an infinite sequence $$ a_1, a_2 , a_3 \cdots $$ And it is given that $$ |a_1| \lt |a_2 -a_1| \lt |a_3 -a_2| \lt |a_4 - a_3| \cdots \\ a_1 \neq 0 $$ (that is the difference between the subsequent terms are increasing and first term cannot be zero)

Can we conclude that the absolute values of subsequent terms are increasing? That is can we conclude $$ |a_1| \lt |a_2| \lt |a_3| \lt |a_4| \cdots $$ Playing around with the inequalities given in the question can give us the information that the alternate terms are increasing (in absolute/numerical value, leaving $a_1$ aside, that is, not comparing $a_1$ with any terms but just caring that it is not zero) but not the consecutive terms. So, I think we cannot conclude that the consecutive terms are numerically increasing.

An explanatory answer is sought.

  • $\begingroup$ You are right. Take $a_1 = 1, a_2 = 3, a_3, = -1$. Now, $|1| < |2| < |4|$ holds but $|a_3| \not > |a_2|$ $\endgroup$ – John White Aug 15 at 6:32
  • 1
    $\begingroup$ @Andrei In that case $|a_i| < |a_{i+1}|$ still holds $\endgroup$ – John White Aug 15 at 6:33
  • $\begingroup$ Thanks @JohnWhite. I will delete my comment $\endgroup$ – Andrei Aug 15 at 6:38
  • $\begingroup$ @JohnWhite Thank you. I think that’s the most we can conclude from that (alternate terms are increasing numerically) much info, right? $\endgroup$ – Knight Aug 15 at 6:41
  • $\begingroup$ That is not true either. The same counterexample above works. $\endgroup$ – John White Aug 15 at 6:42

Consider the sequence $b_n:=c_n(1-\tfrac{1}{n})$ where $c_n\in\{+1,-1\}$, and the sequence $a_n:=\sum_{i=1}^nb_i$.

Then $|a_{n+1}-a_n|=|b_{n+1}|=1-\tfrac{1}{n+1}$ is increasing, yet because of the random choice of $c_i$ there is no telling whether $a_n$ is increasing or decreasing. Here is an example generated by a random choice of $c_n$.

enter image description here

| cite | improve this answer | |
  • $\begingroup$ But in your example I cannot see how alternate terms are increasing in absolute values. Can you please make me see it? I really liked your sophisticated example. $\endgroup$ – Knight Aug 15 at 7:35
  • $\begingroup$ Should alternate terms be increasing? It is not stated in the question. $\endgroup$ – Chrystomath Aug 15 at 7:40
  • $\begingroup$ But we can obtain it by doing some algebra with the inequalities. $\endgroup$ – Knight Aug 15 at 7:41
  • $\begingroup$ ? Take $c=(+1,+1,-1,-1)$ in the example, then $a_1$ increases to $a_3$ but decreases to $a_5$. $\endgroup$ – Chrystomath Aug 15 at 7:45
  • 1
    $\begingroup$ Welcome! My mistakes don't help. $\endgroup$ – Chrystomath Aug 15 at 8:04

One counterexample suffices, and you can produce one with just three terms. If you want to go a little further and show that there need not even be a point beyond which the terms increase in absolute value, you have to work a little harder, but not much. For instance, let $a_1=1$, and in general let

$$a_{n+1}=\begin{cases} a_n-n,&\text{if }n\text{ is odd}\\ a_n+n,&\text{if }n\text{ is even,} \end{cases}$$

so that you get the sequence $1,0,2,-1,3,\ldots\;$; it’s not hard to show by induction that in that case $a_{2n-1}=n$ and $a_{2n}=1-n$ for all $n\in\Bbb Z^+$. Evidently $|a_{n+1}-a_n|=n$ for $n\in\Bbb Z^+$, but $|a_{2n}|=n-1<n=|a_{2n-1}|$ for $n\in\Bbb Z^+$.

| cite | improve this answer | |
  • $\begingroup$ Alternate terms are increasing numerically. But subsequent terms are not. Yes, I can see it in your example. $\endgroup$ – Knight Aug 15 at 6:46

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.