# Sequences with multiple limit points

Is it correct to say that the sequence $\left<(-1)^nn\right>$ tends to both $\infty$ and $-\infty$, since it eventually alternately approaches (oscillates between) $\infty$ and $-\infty$?

Or should I instead say that the sequence $\left<(-1)^nn\right>$ tends to neither $\infty$ and $-\infty$, since there is no single limit point (considering for our purposes that $\infty$ is a limit point)?

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Definition: A sequence $\{s_n\}$ tends to $\infty$ if for any value $x$, there exists an $N$, such that for all $n > N$, we have $s_n > x$. Similarly, a sequence $\{s_n\}$ tends to $-\infty$ if for any value $x$, there exists an $N$, such that for all $n > N$, we have $s_n < x$.

You can show from this definition, that a sequence cannot tend to 2 different values in $\mathbb{R} \cup \{-\infty, \infty\}$. Likewise, your sequence $\{ (-1)^n \cdot n\}$ doesn't tend to $\infty$ or $-\infty$.

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+1, couldn't have said it any better. However, I would like to mention that $\left<(-1)^n\right>$ does have two limit points, namely $1$ and $-1$. Note that "tending to $\infty$" in particular means having no limit point. –  Jesko Hüttenhain Jan 9 '13 at 8:35
Thanks for the argument-from-definition reminder! I know I could simply elaborate my point that the sequence alternately approaches $\infty$ and $-\infty$ (by referring to its subsequences for example), but I still find it a tad misleading to claim that the sequence approaches neither $\infty$ nor $-\infty$, even though that is technically correct... –  Ryan Jan 9 '13 at 8:45
@JeskoHüttenhain I like the question "Show that a sequence has no limit points if and only if the sequence tends to $\infty$ or $-\infty$". It has ideas of compactness, which isn't officially introduced. –  Calvin Lin Jan 9 '13 at 8:50