Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Consider sequences $(x_n)_{n=1}^\infty\subset\mathbb R$. Is there a name for the following property?

There exists $L\in\mathbb N$ such that:

$$\lim\limits_{k\rightarrow\infty}x_{(kL+m)}=x^\ast_m$$

for $m\in\{0,1,2,\dots,(L-1)\}$.

Here the $x^\ast_m$'s are not necessarily equal.

As an example, the sequence $x_n=(-1)^n +\frac n{n+1}$ has $x_{2n}\rightarrow 2$ and $x_{2n+1}\rightarrow 0$ as $n\rightarrow\infty$. In this case $L=2$ (choosing $L$ to be minimal).

share|improve this question
    
Basically, you have $L$ sequences "interweaved". I doubt there is any special name for them, but perhaps some analyst will know better than I. –  Arturo Magidin Dec 17 '11 at 5:03
    
If I had naming rights, I would call it an $L$-colored sequence. –  Bruno Joyal Dec 17 '11 at 5:11
2  
I'd call it a limit-periodic sequence, but I'm not aware that a standard term exists for that. –  alex.jordan Dec 17 '11 at 5:18
    
Thank you all for your comments. @alex.jordan, some research following your comment has lead me to believe that such sequences are called "asymptotically periodic". I will post an answer below. –  matt Dec 18 '11 at 2:03

1 Answer 1

up vote 3 down vote accepted

A sequence with the described property is called an asymptotically periodic sequence.

A sequence is asymptotically periodic if its terms approach those of a periodic sequence. That is, the sequence $x_1,x_2,x_3,\dots$ is asymptotically periodic if there exists a periodic sequence $a_1,a_2,a_3,\dots$ for which: $$\lim_{n\to\infty}x_n - a_n = 0$$

Source: http://en.wikipedia.org/wiki/Periodic_sequence#Generalizations

share|improve this answer
    
+1 for the wikipedia reference. But, in analogy to this terminology, one might call a convergent sequence asymptotically constant, which I do not like very much. :-) –  Srivatsan Dec 18 '11 at 2:11

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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