# 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.

## 3 Answers

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, $L=K$.

• the last ineuqality that u used is only true if u had two constants that the substraction of them in absolute value is less than epsilon , then we can deduce that the are equal ? – F1sargyan Nov 20 '15 at 13:34
• I am using that if $c$ is nonnegative and $c < \delta$ for all positive $\delta$, then $c=0$. @F1sargyan, so in our case $|L-K| = 0$ and so $L=K$. – user99914 Nov 20 '15 at 13:37

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 $0$.

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 $K=L$.

• +1 While less formal in terms of proof then John Ma's, this is exactly what I came up with as well... – imranfat Nov 20 '15 at 16:47