Let’s start by recalling an important theorem of real analysis:

THEOREM. A necessary and sufficient condition for the convergence of a real sequence is that it is bounded and has a unique limit point.

As a consequence of the theorem, a sequence having a unique limit point is divergent if it is unbounded. An example of such a sequence is the sequence $$u_n = \frac{n}{2}(1+(-1)^n),$$ whose initial values are $$0, 1, 0, 2, 0, 3, 0, 4, 0, 5, 6, \dots$$ $(u_n)$ is an unbounded sequence whose unique limit point is $0$.

Let’s now look at sequences having more complicated limit points sets.

A sequence whose set of limit points is the set of natural numbers

Consider the sequence $(v_n)$ whose initial terms are $$1, 1, 2, 1, 2, 3, 1, 2, 3, 4, 1, 2, 3, 4, 5, \dots$$ $(v_n)$ is defined as follows $$v_n=\begin{cases} 1 &\text{ for } n= 1\\ n – \frac{k(k+1)}{2} &\text{ for } \frac{k(k+1)}{2} \lt n \le \frac{(k+1)(k+2)}{2} \end{cases}$$ $(v_n)$ is well defined as the sequence $(\frac{k(k+1)}{2})_{k \in \mathbb N}$ is strictly increasing with first term equal to $1$. $(v_n)$ is a sequence of natural numbers. As $\mathbb N$ is a set of isolated points of $\mathbb R$, we have $V \subseteq \mathbb N$, where $V$ is the set of limit points of $(v_n)$. Conversely, let’s take $m \in \mathbb N$. For $k + 1 \ge m$, we have $\frac{k(k+1)}{2} + m \le \frac{(k+1)(k+2)}{2}$, hence $$u_{\frac{k(k+1)}{2} + m} = m$$ which proves that $m$ is a limit point of $(v_n)$. Finally the set of limit points of $(v_n)$ is the set of natural numbers.

A sequence whose set of limit points is the segment $[0,1]$

Taking advantage of the sequence $(v_n)$, let’s now consider $(r_n)$ whose initial terms are $$\frac{1}{2}, \frac{1}{3}, \frac{2}{3}, \frac{1}{4}, \frac{2}{4}, \frac{3}{4}, \frac{1}{5}, \frac{2}{5}, \frac{3}{5}, \frac{4}{5}, \frac{1}{6}, \frac{2}{6}, \frac{3}{6}, \frac{4}{6}, \frac{5}{6}, \dots$$ Formal definition of $(r_n)$ is $$r_n=\begin{cases} \frac{1}{2} &\text{ for } n= 1\\ \frac{n}{k+2} – \frac{k(k+1)}{2(k+2)} &\text{ for } \frac{k(k+1)}{2} \lt n \le \frac{(k+1)(k+2)}{2} \end{cases}$$ The values of the sequence $(r_n)$ are in $(0,1) \cap \mathbb Q$. Moreover, one can notice that $(r_n)$ takes each rational number of $(0,1)$ as value an infinite number of times. Indeed for $\frac{p}{q} \in (0,1)$ with $1 \le p \lt q$ and $m \ge 1$ we have $$\frac{(mq-2)(mq-1)}{2} \lt \frac{(mq-2)(mq-1)}{2} + mp$$ and $$\begin{aligned} \frac{(mq-2)(mq-1)}{2} + mp &\le \frac{(mq-2)(mq-1)}{2} + m(q-1)\\ &\le \frac{(mq-2)(mq-1)}{2} + mq-1\\ &= \frac{(mq-1)mq}{2} \end{aligned}$$ Hence $$\begin{aligned} r_{\frac{(mq-2)(mq-1)}{2} + mp} &= \frac{(mq-2)(mq-1)}{2mq} + \frac{mp}{mq} – \frac{(mq-2)(mq-1)}{2mq}\\ &= \frac{mp}{mq}\\ &= \frac{p}{q} \end{aligned}$$ proving the desired result. As the rational numbers of the segment $(0,1)$ are dense in $[0,1]$, we can conclude that the set of limit points of $(r_n)$ is exactly the interval $[0,1]$.

A sequence whose set of limit points is $\mathbb R$

Consider the real function $$\begin{array}{l|rcll} f : & (0,1) & \longrightarrow & \mathbb R\\ & x & \longmapsto & \frac{2x-1}{x(1-x)} \end{array}$$ One can verify that $f$ is continuous, strictly increasing and $$\lim\limits_{x \to 0^+} f(x) = -\infty, \ \lim\limits_{x \to 1^-} f(x) = +\infty.$$ Therefore $f$ is a bijection from $(0,1)$ onto $\mathbb R$. We claim that the set of limit points of the rational sequence $(f(r_n))$ is $\mathbb R$.

For $y_0 \in \mathbb R$, let’s take the unique $x_0 \in (0,1)$ such that $f(x_0)=y_0$. As the set of limit points of $(r_n)$ is $[0,1]$, one can find a subsequence $(r_{j(n)})$ converging to $x_0$. $f$ being continuous, the sequence $(f(r_{j(n)}))$ converges to $f(x_0)=y_0$, concluding our proof.