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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 $a_n \rightarrow -\infty$?

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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.

(What this condition says is that the differences are eventually either small and positive or negative. The cauchy condition says that the differences are small in absolute value.)

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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 $L=\lim a_n$.

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