# Prove a sequence converges to a finite $L$ or diverges to $-\infty$.

Let $a_n$ a sequence such that: $\forall \varepsilon>0\ \exists N:\ \forall n>m\ge N:a_n-a_m < \varepsilon$.
Prove $a_n$ converges to a finite $L$ or diverges to $-\infty$.

So, I'm reading this proof which going as follow:

We'll prove that $a_n$ diverges to $-\infty$ and that will be sufficient.
If $a_n$ diverges to $-\infty$ then it's not a Cauchy sequence. Let's write its negation:

$$\exists {\varepsilon _0} > 0: \forall N \ \exists\ m,n > N:\left| {{a_n} - {a_m}} \right| \ge {\varepsilon _0}$$

Apparently, I suppose at this point to build an index series such that $a_i$ is decreasing.

Can you help me with that?

By the way, why is it sufficient to show the case of $a_n \rightarrow -\infty$?

Take $\epsilon = 1$. then, there exists $N$ such that for all $n,m \geq N$, $a_n - a_m < 1$. In particular, take $m=N$, $a_n-a_N <1$ or $a_n < 1+a_N$ for all $n \geq N$, so $\lim \sup a_n < 1+ a_N < \infty$. Thus, it cannot diverge to $\infty$. You still need to show that the sequence converges to a finite limit or to $-\infty$. Think of the case when the sequence is cauchy and when its not cauchy to get the divergence.
Let $L=\limsup a_n$. Show that $L<\infty$. If $L=-\infty$, we are done. So assume $L>-\infty$ and show that for any $\epsilon>0$, we have $a_n>L-\epsilon$ for almost all $n$. Conclude that $L=\lim a_n$.