A bounded interval $I\subset{\mathbb R}$, say $I:=[{-1},1]$, contains all real numbers $x$ with $-1\leq x\leq1$, which is a huge number of points.
If you are given a ${\it set}\ A$ of "special points" in this interval, say $A:=\{x\in I\>|\>e^{2\pi i/x}=1\}$, then such a set may have accumulation points. An ${\it accumulation\ point\ \xi\ of\ the\ set}\ A$ is defined by the property that any punctuated neighborhood $\dot U$ of $\xi$ contains at least one point from $A$. If $A$ is an infinite bounded set then by Bolzano's theorem there is at least one such accumulation point which may or may not belong to $A$.
Sometimes the given set $A$ has a natural numbering, i.e., there is a bijective function $\phi:\ {\mathbb N}\to A, \ n\mapsto a_n$ which "produces" the points of $A$ one for one. The accumulation points of the set $A$ are also accumulation points of the sequence $(a_n)_{n\geq 0}\ $ (see below). A point $\xi$ is called an ${\it accumulation\ point\ of\ the\ sequence}\ (x_n)_{n\geq0}$ if any neighborhood $U$ of $\xi$ contains $x_n$'s with arbitrary large $n$. But note that in a plot of the situation you might see nothing special: The sequence defined by $x_n:=(-1)^n$ has the two accumulation points $\pm1$ which is all you can see.
Now comes the question you raised about the situation when a given set $A$ has exactly one accumulation point $\xi$. Such a set is countable to begin with (I omit the proof), therefore this set can be bijectively produced by a sequence $(x_n)_{n\geq0}$. Maybe this set was defined as the image set of such a sequence to begin with. I claim that for any such sequence $\lim_{n\to\infty} x_n=\xi$.
Proof: Consider an $\epsilon>0$. If there were infinitely many $x_n\geq\xi+\epsilon$ then these $x_n$ would form an infinite bounded set without an accumulation point, and similarly it is impossible that infinitely many $x_n$ are $\leq\xi-\epsilon$. It follows that there is an $n_0$ with $x_n\in U_\epsilon(\xi)$ for all $n>n_0$.
