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Assume $(a_n)_{n=1}^\infty$ has the following property: For all $\epsilon>0$ exists some $N\in\Bbb N$ such that for every $n>m\ge N$, $a_n-a_m<\epsilon$. We want to prove that $a_n$ converges to a real limit or to $-\infty$.

This condition implies that $a_n$ is bounded above. So, I tried to show what happens if it's bounded below or unbounded below. When bounded below, by Bolzano-Weierstrass $a_n$ has some convergent subsequence, $a_{n_k} \rightarrow L$, so for all $k\ge K$ for some $K\in \Bbb N$, $|a_{n_k}-L|<\epsilon \iff -\epsilon<a_{n_k}-L<\epsilon$ .

I tried to use this to prove the convergence of the entire sequence but it didn't work - I tried to take some $n>n_k>max(N, n_K)$, $a_n-L=a_n-a_{n_k}+a_{n_k}-L <2\epsilon$, but now I can't get the left inequality right. This makes me question my entire process. Maybe I should've gone about this a different way?

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Your approach is fine.

If $(a_n)_{n\in\mathbb N}$ is undounded below, Take $M<0$. The inequality $a_n<M-1$ holds infinitely often and there is a $N\in\mathbb N$ such that $$n>m\geqslant N\implies a_n-a_m<1.$$Therefore, if $n\gg1$, $a_n<(M-1)+1=M$. This proves that $\lim_{n\to\infty}a_n=-\infty$.

Otherwise, let $L=\liminf_na_n$. Take $\varepsilon>0$. Then the inequality $|a_n-L|<\frac\varepsilon2$ hold infinitely often and $a_n>L-\frac\varepsilon2$ if $n\gg1$. Now, take $N\in\mathbb N$ such that$$n>m\geqslant N\implies a_n-a_m<\frac\varepsilon2.$$Then, if $n\gg1$, $|L-a_n|<\varepsilon$.

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    $\begingroup$ I don't understand the sentence beginning with "Therefore". $\endgroup$ – DanielWainfleet Jan 22 '18 at 1:19
  • $\begingroup$ @DanielWainfleet I suppose that that's due to a typo in the previous sentence: I wrote “$a_n<M-$”, whereas I should have written “$a_n<M-1$”. The answer has already been edited. $\endgroup$ – José Carlos Santos Jan 22 '18 at 8:45
  • $\begingroup$ Good A...................... $\endgroup$ – DanielWainfleet Jan 24 '18 at 5:04
  • $\begingroup$ Not sure I understand. How does the fact that $a_n<M-1$ for an infinite number is $n$'s, imply that for all $1<<m<n$, $a_n<M$? This is unclear to me as the first inequality does not necessarily hold for all $m$'s in the range. $\endgroup$ – Noa Jan 24 '18 at 8:26
  • $\begingroup$ @Noa Take $N$ large enough so that $n>m\geqslant M\implies a_n-a_m<1$. Take $m\geqslant N$ such that $a_m<M-1$. Then,$$n\geqslant m\implies a_n=a_n-a_m+a_m<1+M-1=M.$$ $\endgroup$ – José Carlos Santos Jan 24 '18 at 8:31

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