I've already done a lot of searching around MSE for this. (In particular, If a sequence has two convergent subsequences with different limits, then it does not converge.)
The Cauchy criterion has not been covered yet.
I would like to show – without using the fact that a converging sequence has converging subsequences, in which both cases have the same limits – that if I have two subsequences with differing limits, the sequence does not converge.
That is, $\exists \epsilon > 0$ such that for all $M \in \mathbb{N}$, there is an $n \geq M$ such that $|x_n – x| \geq \epsilon$.
(I hope my negation above is correct.)
I am given two subsequences of $x_n$, namely $x_{n_i} \overset{i \to \infty}{\to}a$ and $x_{m_i} \overset{i \to \infty}{\to}b$, $a \neq b$.
Suppose, by way of contradiction, that $\lim\limits_{n \to \infty}x_n = x$.
I have the subsequence convergence, so something like $|x_{n_i} – a| < \text{?}$ and $|x_{m_i} – b| < \text{??}$.
I thought maybe to use a trick here:
$$|x_n – x| = |x_{n}-x_{n_i}+x_{n_i}-a-x_{m_i}+x_{m_i}-b+b+a-x|$$
but I'm not sure how to proceed from here (triangle inequality isn't helpful).
Best Answer
As you know, the definition of $a_n \to c$ is $$\forall \epsilon>0\text{ } \exists N \in \Bbb{N} \text{ such that } s > N \implies |c - x_s| < \epsilon$$ The negation of this definition is simply, for any $c$, $$\exists\epsilon>0\text{ such that } \forall N \in \Bbb{N} \text{ } \exists s > N \text{ such that } |c - x_s| \ge \epsilon$$ And now we merely wish to show that this second statement applies to the series in question. Pick a value of $c$. We wish to show that the sequence does not converge to $c$. Now, either $b \ne c$ or $a \ne c$ (or both). Without loss of generality, assume $a \ne c$. Then let $\epsilon = |a-c|/2$. Now let $N$ be any natural number.
Our task, now, is to find some $s>N$ so that $|c - x_s| \ge \epsilon$. Since $(x_{n_i})$ converges to $a$, let $s = n_i$, where $n_i$ is large enough so that $|a-x_{n_i}| < \epsilon$ and $n_i > N$. Then $$|c - a| = |c - x_s + x_s - a| \le |c - x_s| + |a - x_s| < |c - x_s| + \epsilon = |c - x_s| + |c-a|/2$$ Therefore, $$|c - x_s| > |c-a|/2 = \epsilon$$ And we're done. We have shown that for any real number $c$, our sequence cannot converge to $c$. I hope this was helpful.