Prove that: $\lim\limits_{n \to \infty} b_n = \lim\limits_{n \to \infty} a_n$. It's given that $a_n$ and $b_n$ two convergent sequences. prove that if for each $even$ $n$: $a_n \le b_n$ and for each $odd$ $n:$ $b_n  \le a_n$ then $\lim\limits_{n \to \infty} b_n = \lim\limits_{n \to \infty} a_n$.
SOLUTION: I tried to use the definition of limit.
that for each $\epsilon>0$ there is such $N_1$ s.t. for each $n>N_1$: $L-\epsilon \lt a_n < L+\epsilon$
and  for each $\epsilon>0$ there is such $N_2$ s.t. for each $n>N_2$: $K-\epsilon \lt b_n < K+\epsilon$
and I thought about the idea that Odd and Even numbers together cover all the numbers. and I treid to prove by contradiction, that $L \neq K$ and try to reach a place where K must be euqal to L. but didn't quite know how to do it.
any kind of help would be appreciated. 
 A: Differences of convegent sequences are convergent and subsequences of convergent sequences have the same limit. Then your assumptions imply $\lim_n(a_{2n}-b_{2n}) \le 0$ and $\lim_n (a_{2n+1}-b_{2n+1}) \ge 0$. But each of them is equal to $\lim_n(a_n-b_n)$, hence it is $0$.
A: Let $N = \max\{N_1, N_2\}$, $n_1, n_2 \ge N$, $n_1$ odd and $n_2$ even. Then 
$$\begin{split}
 K -\epsilon < b_{n_1} \le a_{n_1} < L +\epsilon &\Rightarrow K-L <2\epsilon, \\
 L -\epsilon < a_{n_2} \le b_{n_2} < K +\epsilon &\Rightarrow L-K <2\epsilon.
\end{split}$$
So $|L-K| <2\epsilon$. As $\epsilon >0$ is arbitrary, $L=K$. 
A: Consider the sequences  $(c_n)_n$, $(d_n)_n$, $(x_n)_n$, $(y_n)_n$ defined by 
$$ c_n=a_{2n}$$ $$ d_n=b_{2n}$$ $$ x_n=a_{2n+1}$$ $$ y_n=a_{2n+1}$$  Then Clearly these sequences are convergent and we have  $$\lim_{n \rightarrow  \infty } c_n= L $$  $$\lim_{n \rightarrow  \infty } d_n= K $$  $$\lim_{n \rightarrow  \infty } x_n= L $$  $$\lim_{n \rightarrow  \infty } y_n= K $$
Note that for any $n$, we have  $c_n \leq  d_n$ , hence taking limits on both sides we get  $ L\leq  K$. On the other hand $ y_n \leq x_n $, so also by taking limits we get  $K \leq  L$. Thus  $K=L$.
