Given a sequence {bn}, with two subsequence which both converge, proof that {bn} not have to be convergent Given a sequence $\{bn\}$, with two subsequences which both converge, prove that $b_n$ need to  converge. 
Given $bn$ and subsequences $b_{2n}$ and $b_{2n+1},$
where $b_{2n}$ and $b_{2n+1}$ are both covergent,
Show that $bn$ dosen´t have to be convergent.
I know that if $bn$ is covergent then all of the subsequences of it must have the same limit point let´s say $x$.
But if $b_{2n}$ has limit point $y$ and $b_{2n+1}$ has limit point $z$ then $b_n$ cannot be convergent, but how can i  show that?? 

Take $b_n=(-1)^n$.
$\{b_{2n}\}$ converges and $\{b_{2n-1}\}$ converges, but $\{b_n\}$ does not.
Let $A=\lim\limits_{n\rightarrow+\infty}b_n$ and $\epsilon=\frac{1}{2}.$
Thus, for all $N>0$ there is $n>N$ for which $|b_n-A|\geq\epsilon$.
