A sequence converges if and only if it has exactly one limit point - How come is this true? I am following Nocedal's Numerical Optimization book; in the Appendix about Analysis and Topology I came accross the highlighted argument:

A sequence converges implies it has one limit point: This is OK, I can show this by contradiction: Assume an arbitrary limit point $x'$ which is different from the converged point $x$ and show that the both cannot coexist.
But the other way around of the equivalence seems hardly true to me: A sequence has exactly one limit implies it converges.
For example, if we think about a sequence which has for every odd $i$, $1/i$ and for every even $i$, $2i+1$, then it has only one limit point, which is $0$ but it is obvious that it does not converge.
Am I missing something here?
 A: You're missing that infinities should be included as possible limit points. 
Then your sequence has 2 limit points: $0$ and $+\infty$. 
A: A proof goes like this.
proof
We exploit the following very much important property of compact sets:

Any sequence of a compact metric space has a convergent subsequence.

(Most of the corresponding contexts reserve this for the definition of compactness though compactness can be defined in another ways but the result is the same)
Since $\{x_k\}\in \Bbb R^n$ eventually falls in $\{x\ \ |\ \ |x-r|\le \epsilon\}$ ( because of the definition of limit point) for any $\epsilon>0$ where $r$ is the limit point of $x_n$ and $\{x\ \ |\ \ |x-r|\le\epsilon\}$ is compact , then $x_k$ has a convergent subsequence namely $a_n$. Removing it from $x_n$, we can derive another convergent subsequence namely $b_n$. We can follow this procedure infinitely many times. Now let $x_n$ diverge. Then it must have at least two subsequences convergent to two different values (if not i.e. if all the convergent subsequences tend to same number, then so will do the sequence itself). Then we must have at least two limit points which is obviously a contradiction and the proof finishes up.
