How does a sequence who does not have a limit point behave? In $R$ every bounded sequence have a limit point. Suppose the sequence and bounded and further say that it does not have a limit point. Would this directly conclude that it goes to infinity?
Can't there be sequences who go to infinity but have limit points? 
I mean is the statement below true:
A sequence diverges to infinity if and only if it doesn't have any limit points.
 A: Assume the sequence $u_n$ has a limit point $a$ and goes to $+\infty$.
take $\epsilon=1$.
$\lim_{n\to +\infty}u_n=+\infty \implies
\exists n_0 \in \mathbb N :$
$\forall n\geq n_0  \;  u_n\geq a+2$
so,  $(a-\epsilon,a+\epsilon)$ is empty . This is in contradiction with the hypothesis $a$ is a limit point.
we can have a unbounded sequence which has a limit point.
$u_n=n$ if $n$ is even and
$u_n=0$ if $n$ is odd.
A sequence which is not bounded doesn't go automatically to $\infty$.
A: Let $(x_n)\subset\mathbb R$ be a sequence of real numbers. Then exactly one of the following two statements is true.


*

*$\lim_{n\to\infty}|x_n|=\infty$

*There exists a subsequence $(x_{n_k})$ of $(x_n)$ that converges to some (finite) real number (i.e. $(x_n)$ has a limit/accumulation point).


If a sequence does not have any finite limit points (so (2) is not satisfied), then this does not mean that $\lim_{n\to\infty}x_n=+\infty$ or $\lim_{n\to\infty} x_n=-\infty$. For instance, 
$$
x_n=\begin{cases}n, & n \text{ odd} \\ -n, & n \text{ even}\end{cases}=(1,-2,3,-4,5,-6,...).
$$
Conversely, if we know that $\lim_{n\to\infty}x_n=+\infty$ or $\lim_{n\to\infty}x_n=-\infty$, then $\lim_{n\to\infty}|x_n|=\infty$, and in this case $(x_n)$ cannot have any finite limit points.
A: One of the corollary of the Bolzano Weierstrass Theorem is that any real sequence has at least one limit point in $\Bbb R \cup \{ +\infty, -\infty\}$. Hope this answers your question.
Indeed let $u_n$ be a real sequence. If $u_n$ is bounded the theorem states that one of its subsequence converges to a real.
If it is not bounded, that means :
$\forall M \in \Bbb R, \exists n \in \Bbb N, u_n \ge M$ or $\forall M \in \Bbb R, \exists n \in \Bbb N, u_n \le M$
You can therefore construct by induction a subsequence of $u_n$ by selecting an extraction $\phi$ such that $u_{\phi(n+1)} \le \min \{u_k, k \le \phi(n)\}-1$ (or $u_{\phi(n+1)} \ge \max \{u_k, k \le \phi(n)\}+1$), which then diverges either to $+\infty$ or $-\infty$
Thus you can deduce from this that any sequence that does not have a limit point diverges to one (or both) of the infinities.
