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Prove that :

If lim inf$s_n$ = lim sup$s_n$ , then lim $s_n$ is defined and lim $s_n$ = lim inf$s_n$ = lim sup$s_n$

case(1) lim inf$s_n$ = lim sup$s_n$ = +∞ then lim $s_n$ = +∞.

case (2) if lim inf$s_n$ = lim sup$s_n$ = −∞ then lim $s_n$ = −∞. case(3)

case(3) Suppose that lim inf $s_n$ = lim sup $s_n$ = $s$ where $s$ is a real number. We need to prove lim $s_n$ = $s$. Let $ε > 0$ Since $s$ = lim vN there exists a positive integer $N_0$ such that $|s−sup{s_n :n>N_0}|$<ε. Thus sup${s_n :n>N_0}$<$s+ε$ , so $s_n$ <$s+ε$ for all n>$N_0$. (1) Similarly, there exists $N_1$ such that $|s − inf {s_n : n > N_1 }| $ < ε, hence inf${s_n:n>N_1}$ >$s−ε$, hence $s_n$ >$s−ε$ forall $n>N_1$. (2) From (1) and (2) we conclude $s−ε$<$s_n$<$s+ε $ for n>max${N_0,N_1}$ , equivalently $|s_n − s| $ < $ε$ for n > max${N_0,N_1}$. This proves lim $s_n$ = $s$ as desired.

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    $\begingroup$ Unreadable. Please format with $\LaTeX$. | We need to know your effort and/or mathematics level so we can write an answer properly. $\endgroup$ – user202729 Nov 13 '17 at 11:05
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    $\begingroup$ Make explicit the points you don't understand. $\endgroup$ – Jean Marie Nov 13 '17 at 11:25
  • $\begingroup$ My edit was to delete one dollar-sign. It would be better to put dollar-signs only before the very beginning and after the very end of each formula. Use \sup and \max and \lim and \infty between dollar-signs. And don't omit the space immediately after each of them. $\endgroup$ – DanielWainfleet Nov 13 '17 at 15:41
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Hint

Indefinite limits can be vaguely defined using a phrase "For however large $K$ there exists $\varepsilon$ such that any values of $s_n$ are higher than $K$.", hence no need to consider supremum for case (1) and infimum for case (2).

I hope I did not spoil the fun playing around with the $\varepsilon$'s, it is a great exercise.

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