Convergence of sequence points to the point of accumulation Wikipedia says:

Every finite interval or bounded interval that contains an infinite number of points must have at least one point of accumulation. If a bounded interval contains an infinite number of points and only one point of accumulation, then the sequence of points converge to the point of accumulation. [1]

How could i imagine a bounded interval of infinite points having a single limit point and that sequence of interval's points converge to that limit point.
Any examples?
[1] http://en.wikipedia.org/wiki/Limit_point
Thank you.

What about: $$\left \{\frac{1}{n}\Bigg| n\in\mathbb Z^+ \right \}\cup\{0\}$$
This set (or sequence $a_0=0,a_n=\frac{1}{n}$) is bounded in $[0,1]$ and has only limit point, namely $0$.
A slightly more complicated example would be:
$$\left \{\frac{(-1)^n}{n}\Bigg| n\in\mathbb Z^+ \right \}\cup\{0\}$$
In the interval $[-1,1]$. Again the only limit point is $0$, but this time we're converging to it from "both sides".
I believe that your confusion arises from misreading the text. The claim is that if within a bounded interval (in these examples, we can take $[-1,1]$) our set (or sequence) have infinitely many points and only one accumulation point, then it is the limit of the sequence.
There is no claim that the interval itself has only this sequence, or only one limit/accumulation point.
