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.

Define a sequence for $\displaystyle (a_n)^\infty_{n=1}$ such that $\displaystyle \lim_{n\to\infty} a_{n^2}$ exists but $\displaystyle \lim_{n\to\infty} a_n$ does not exists.


My first intuition is if we are not finding a sepecifc element or sequence. What makes the difference.

Since $(n=1,2,...,n-1,n)$ and $(n=1,2,...,n,...,n^2-1,n^2)$ they both will go to $\infty$

or the question actually means $(n=1,4,9,16,...,(n-1)^2,n^2$ ??

since both limits are using the same sequence so $a_{n^2}$ must be a subsequence of $a_n$ and when $n\to\infty$, what makes the difference??

Thanks!!

share|improve this question
    
The notation suggests, that $a_{n^2}$ is meant to be $a_1,\ a_4,\ a_9,...$. –  sonystarmap Feb 4 '13 at 20:34
    
What about a constant subsequence if the index is a square and the index itself if not. –  Sigur Feb 4 '13 at 20:34
2  
Yes I agree with Sigur. Somehting like $a_n = 0$ for $n=m^2$ for some $m$ and $a_n=n$ else. –  sonystarmap Feb 4 '13 at 20:35
1  
Make $a_k$ say $1$ if $k$ is a perfect square, and $0$ if it is not. –  André Nicolas Feb 4 '13 at 20:39
    
@Paul It's true that if $(a_n)$ goes to $\infty$ then so does $(a_{n^2})$. However, the question doesn't require that $a_n \to \infty$: there are plenty of other ways for a sequence to not have a limit. –  Erick Wong Feb 4 '13 at 20:58

1 Answer 1

up vote 4 down vote accepted

Consider the sequence $(a_k)$ given by $$a_k=\begin{cases} 1, & k=n^2 \text{ for some natural $n$}, \\ k, & k\neq n^2 \text{ for any natural $n$}. \end{cases}$$

share|improve this answer

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.