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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??


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The notation suggests, that $a_{n^2}$ is meant to be $a_1,\ a_4,\ a_9,...$. – cmmndy 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
Yes I agree with Sigur. Somehting like $a_n = 0$ for $n=m^2$ for some $m$ and $a_n=n$ else. – cmmndy Feb 4 '13 at 20:35
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 $$a_k=\begin{cases} 1, & k=n^2 \text{ for some natural $n$}, \\ k, & k\neq n^2 \text{ for any natural $n$}. \end{cases}$$

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