Prove that if $({a_n} )$ is a sequence with two convergent subsequences $\{a_{n_k}\}$ and $\{a_{m_k}\}$ 
Prove that if $({a_n}
)$ is a sequence with two convergent subsequences $\{a_{n_k}\}$ and $\{a_{m_k}\}$ such that $\lim_{k\to\infty}\{a_{n_k}\} \neq \lim_{k\to\infty}\{a_{m_k}\}$, then $({a_n})$ does not converge.


*

*I am not sure if I am approaching the question correctly but, I have two limits set up to two different answers, $|\{a_{n_k}\} - L|<\epsilon/2$ and $|\{a_{m_k}\} - M|<\epsilon/2$ because I thought if I could somehow show these can not be manipulated to equal $\epsilon$ then their limits could not be equal to each other and therefore $(a_n)$ would not converge. I'm not sure if I can set the limits equal to an answer without proving something first or if I am actually going the right way about this. I am a little confused by the question in general, any help would be much appreciated.

 A: Recall that a sequence $\{a_n:n\in\mathbb{N}\}$ is Cauchy if for any $\varepsilon>0$, there is $N$ such that $n,m\geq N$ then $|a_n-a_m|<\varepsilon$.
We have the following well known result:
Proposition: If a sequence $\{a_n:n\in\mathbb{N}\}$ is convergent, then it is Cauchy.
Here is a short proof: Suppose $a_n\xrightarrow{n\rightarrow\infty}a$. Then, for any $\varepsilon>0$, there is $N$ such that $n\geq N$ implies $|a_n-a|<\varepsilon/2$. Consequently, if $n,m\geq N$,
$|a_n-a_m|\leq |a_n-a|+|a-a_m|<\varepsilon$.
Notice that the negation of being Cauchy is:
$\{a_n\}$ is not Cauchy if there is $\varepsilon>0$ such that for any $N$, there exist $m,m\geq N$ such that $|a_n-a_m|\geq\varepsilon$.

Under the condition of the original OP, we show $\{a_n\}$ is not Cauchy:
Suppose $a_{n_k}\xrightarrow{k\rightarrow\infty}L$, $a_{m_k}\xrightarrow{k\rightarrow\infty}M$ and $M\neq L$. Let $$\varepsilon:=\frac{|L-M|}{4}$$
By assumption, there is $K_0$ such the $k\geq K_0$ implies that
$$|a_{n_k}-L|<\varepsilon$$
and
$$|a_{m_k}-M|<\varepsilon$$
Now, given any integer $N$, choose $k\geq \max(K_0,N)$. Then, $n_k\geq k\geq N$ and $m_k\geq k\geq N$, and
$$
|a_{n_k}-a_{m_k}|\geq \varepsilon
$$
(without loss of generality assume $L<M$ and draw a picture to see what is going on. A rigorous argument is based on the triangle inequality).
This means that $\{a_n\}$ is an a Cauchy sequence and so, $\{a_n\}$ is not convergent.
A: Just assume wlog $L>M$ and $\varepsilon \le \frac{L-M}{3}$ then since we can find $n_0$ such that for any $n\ge n_0$
$$|a^k_n - L|<\varepsilon \quad \land \quad |a^k_m - M|<\varepsilon$$
which are disjointed intervals
$$\frac{-L+4M}{3}<a^k_m<\frac{L+2M}{3}<\frac{2L+M}{3}<a^k_n <\frac{4L-M}{3} $$
then $a_n$ doesn't converge.
