Convergent Subsequences and the Limit of a Series

[Doom of Doom]

Homework Statement

Consider the sequence $$\left\{ x_{n} \right\}$$.

Then $$x_{n}$$ is convergent and $$\lim x_{n}=a$$ if and only if, for every non-trivial convergent subsequence, $$x_{n_{i}}$$, of $$x_{n}$$, $$\lim x_{n_{i}}=a$$.

Homework Equations

The definition of the limit of a series:
$$\lim {x_{n}} = a \Leftrightarrow$$ for every $$\epsilon > 0$$, there exists $$N \in \mathbb{N}$$ such that for every $$n>N$$, $$\left| x_{n} - a \right| < \epsilon$$.

The Attempt at a Solution

Ok, so I easily see how to show that it $$\lim {x_{n}} = a$$, then every convergent subsequence must also converge to $$a$$.
But I'm stuck on how to show the other way.

 

[Dick]

I would say, well isn't a_n a subsequence of itself? But you also said 'non-trivial'. I'm not sure exactly what that means, but can't you split a_n into two 'non-trivial' subsequences, which then converge, but when put together make all of a_n?

 

[Doom of Doom]

Yeah, I asked my prof about this one. To him, apparently "non-trivial" just means that the subsequence is not equal to the original sequence. I don't think it actually has any bearing on the problem.

The trick, he said, is that you have to consider every non-trivial (convergent) subsequence.

I'm not sure I know what that means.

 

[Dick]

Ok, then suppose a_n has two convergent subsequences with different limits. Then does a_n have a limit?