# Limit point versus limit and implications for convergence

I understand that similar questions to this have been asked here (1) and here (2) but I believe my question is different because with regards to (1) I am not asking about the set $X$ that contains the points of my sequence and with regard to (2) because I have the distinct definitions of limit and limit point in mind. That said:

My Question: My textbook, Fundamental Ideas of Analysis by Michael Reed, defines a limit point as:

Definition: Let $(a_n)$ be a sequence of real numbers. A number $d$ is called a limit point if, given $\epsilon > 0$ and an integer $N$, there exists $\color{red}{\mathrm{ an }} \ n \ge N$ so that $|a_n - d| \le \epsilon$.

The definition of a limit is as follows:

Definition: We say that a sequence $(a_n)$ converges to a limit $a$ if, for every $\epsilon \ge 0$, there is an integer $N(\epsilon)$ so that $|a_n - a| \le \epsilon \color{red}{\text{ for all }} n \ge N$.

From these definitions, and given this question:

If a sequence converges it has exactly one limit point. Is the converse true? If not, provide a counterexample

Can I conclude that the converse is indeed true simply by appealing to the definitions, or do I need a tighter proof. And if I do, I'm a little stuck on how to refine this proof.

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Could an unbounded sequence have exactly one limit point? – David Mitra Aug 15 '12 at 16:52
Does a sequence of this form "count"?: $(a_n) = 1$ for $n = 2k, k \in \mathbb N$, $(a_n) = n$ for $n = 2k - 1, k \in \mathbb N$? – Moderat Aug 15 '12 at 16:54
Yes, its only limit point is $1$ but it does not converge. – David Mitra Aug 15 '12 at 16:58
So that sequence I have constructed works as a counterexample then. Thanks @DavidMitra – Moderat Aug 15 '12 at 17:00

The sequence $2,1/2,3,1/3,4,1/4 \cdots, n,1/n \cdots$ has the only limit point $0$, but not converges.
$(1 + (-1)^n).n$
To be explicit: it is the sequence $0,4,0,8,0,12,0,\dots$ – Ben Millwood Aug 15 '12 at 17:22