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The question is:

Let $s_n$ and $t_n$ with $ n \ge 1 $ be two bounded sequences. Assume that there exists a natural number $N$ such that for every $ n \ge N $, we have $ s_n \le t_n $.

Prove that $$ \lim_{n\to\infty} \sup s_n \le \lim_{n\to\infty} \sup t_n. $$

I think its obvious, but the part of proving it is kinda tricky. Can someone please help?

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    $\begingroup$ If it is kinda tricky, then it is not obvious. $\endgroup$ – Andrés E. Caicedo Nov 30 '13 at 22:10
  • $\begingroup$ i understand that but the OP needs to show some type of attempt. $\endgroup$ – Ahaan S. Rungta Nov 30 '13 at 22:10
  • $\begingroup$ Start with the definition of $\limsup$, and work your way from there. There are several different definitions of $\limsup$, but I think they will all work fairly easily. $\endgroup$ – Stephen Montgomery-Smith Nov 30 '13 at 22:12
  • $\begingroup$ What is the definition of $\limsup$ that you're given? Also, it seems the word is "intuitive" rather than "obvious". However, as you progress within mathematics, you'll find that some "intuitively obvious" statements turn out to be false. $\endgroup$ – Omnomnomnom Nov 30 '13 at 22:17
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One definition of $\limsup$ of a sequence $\{ r_{k} \}$ is as follows. Define $$ R_{n} = \sup \{ r_{n}, r_{n+1}, r_{n+2},\ldots \}. $$ Whether finite or infinite, $$ R_{1} \ge R_{2} \ge R_{3} \ge .... $$ converges downward to some finite or infinite limit defined as $\limsup_{n}r_{n}$. For and $n \ge N$, the least upper bound $T_{n}$ of $\{ t_{n},t_{n+1},t_{n+2},\ldots\}$ is an upper bound for $\{ s_{n},s_{n+1},s_{n+2}\}$ because $s_{k} \le t_{k}$ for $k \ge n \ge N$. Therefore the least upper bound $S_{n}$ of $\{ s_{n},s_{n+1},s_{n+2},\ldots\}$ must satisfy $S_{n} \le T_{n}$. So the argument is now reduced to the case of monotone sequences: $$ \limsup_{n} s_{n} = \lim_{n} S_{n} \le \lim_{n} T_{n} = \limsup_{n} t_{n}. $$ The infinite cases can be dealt with directly. If $\limsup_{n} t_{n}=\infty$, there is nothing to show. If $\limsup_{n} s_{n}=-\infty$, there is nothing to prove. In all other cases, both $\limsup$ terms are finite, and $\{ S_{n} \}$, $\{ T_{n} \}$ are non-decreasing, convergent sequences with $S_{n} \le T_{n}$ for $n \ge N$, a case which is assumed to be known.

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  • $\begingroup$ CORRECTION: The last sentence should have read "non-increasing" instead of "non-decreasing". $\endgroup$ – DisintegratingByParts Nov 30 '13 at 23:31
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On the contrary let $S = \limsup s_{n} > \limsup t_{n} = T$. Then by definition of $\limsup$ we have the following:

1a) For any $\epsilon > 0$ there are infinitely many values of $n$ for which $s_{n} > S - \epsilon$.

1b) For any $\epsilon > 0$ we have $s_{n} < S + \epsilon$ for all sufficiently large values of $n$.

2a) For any $\epsilon > 0$ there are infinitely many values of $n$ for which $t_{n} > T - \epsilon$.

2b) For any $\epsilon > 0$ we have $t_{n} < T + \epsilon$ for all sufficiently large values of $n$.

Put $2\epsilon = S - T > 0$ so that $T + \epsilon = S - \epsilon$. Now for all sufficiently large values of $n$ we have $t_{n} < T + \epsilon = S - \epsilon$ and hence $s_{n} \leq t_{n} < S - \epsilon$. This contradicts the fact that $s_{n} > S - \epsilon$ for infinitely many values of $n$. Hence we must have $S \leq T$.

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I'll use the following property:
For a sequence $(a_n)_{n\in\mathbb N}$ the $\limsup a_n$ is the largest limit point of $(a_n)_{n\in\mathbb N}$, where $a$ is a limit point if there is a subsequence $(a_{k_n})_{n\in\mathbb N}$ such that $\lim_{n\to\infty} a_{k_n}=a$.

Assume that $\limsup s_n>\limsup t_n$.
Then there is a subsequence $(s_{k_n})_{n\in\mathbb N}$ of $(s_{n})_{n\in\mathbb N}$ such that $\lim_{n\to \infty} s_{k_n}=\limsup s_n>\limsup t_n$.
Therefore $s_{k_n}>\limsup t_n$ for all large $n$ so $t_{k_n}>\limsup t_n$ for all large $n$.
Now find a subsequence of $t_{k_n}$ that converges ($t_{k_n}$ is bounded) to derive a condradiction.

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