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?? 
 A: 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 $|b_n-A|\geq\epsilon$.
A: You don't have to prove anything because you have the proposition which states that if a sequence has a limit say $x$ then every subsequence must converge to $x$.
The convergence of the subsequences to the same limit is a necessary condition.
If exist  subsequences of $x_n$ which converge to different limits then $x_n$ cannot be convergent
Let  $k$ a non negative integer.
Then all $n \in \mathbb{N}$ can be written: $$n=ks+m$$ where $o \leq m \leq k$(think of this as the euclideian division where a  natural number depends on the remainder $m$).
For instance if you take the number $5$ then each natural number $n$ can be expressed in the form $n=5s+m$  for some $s \in \mathbb{N}$ and $0 \leq m \leq 4$
You can do this for every number $k$ in other words you can partition the set of natural numbers to $k$ disjoint sets.Take for instance $k=10$ and think it your self what are all the possible forms the natural numbers will have.
Now why do i mention this?
If you take a sequence $x_n$ and you prove for instance that the subsequences $x_{3n+1},x_{3n+2},x_{3n}$ (where you chose $k=3$) converge to the same limit $l$ then $x_n$ converges to $l$ .
You can do this for every $k$(partition)
Your case is when $k=2$ where we partition the set of natural numbers to two disjointsets, the sets of odd numbers and the set of event numbers.
According to your question if you prove that $x_{2n},x_{6n},x_{3n}$ converge to the same limit then you don't prove that $x_n$ has the same limit.
I hope this helps.
