Accumulation points of a sequence I was wondering , if a sequence converges does that mean that the sequence have exactly one accumulation point that occurs at it's limit , and if a sequence doesn't converge then it doesn't have any accumulation point ? is that true !
 A: If it converges it has only one accumulation point.  If it doesn't converge it may have any number of accumulation points. Take (1,-1,1,-1,1,-1,..) for an example with 2 accumulation points.  If we want distinct points accumulating use:
$   a(n) = \left\{
     \begin{array}{lr}
       \frac{1}{n}    & : n\equiv0 (2) \\
       1-\frac{1}{n}  & : n\equiv1(2) \\
     \end{array}
   \right.
$
To see that a convergent sequence has exactly one accumulation point:
Your sequence is assumed to converge $a_n \rightarrow c$. Suppose there were two distinct points of accumulation $c,d$.  Then there is some positive distance between $c$ and $d$, $|c-d|=\epsilon >0 $.  By convergence there is some $N$ such that $n \geq N \implies |a_n-c|< \frac{\epsilon}{2}$.  But then the sequence is always more than $\frac{\epsilon}{2}$ away from $d$ for $n \geq N$ so $d$ couldn't have been an accumulation point.
A: If $x$ is an accumulation point of $x_n$ means that every punctured neighborhood of $x$ contains some $x_i$ then neither statement is correct.  
For the first statement, note the constant sequence converges yet has no accumulation points.  
For the second statement, note a sequence which does not converge can still have accumulation points.  For example any enumeration of the rationals has every real number as an accumulation point.  
