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Q.1 Assume $s_n\ge a$ for all but finitely many n, then $\lim s_n\ge a$.

I think using contradiction can prove it but i wonder if there is a more direct proof.

Q.2 suppose that there exists $N_0$ such that $s_n\le t_n$for all $n>N_0$, prove that if $\lim s_n$ and $\lim t_n$ exist, then $\lim s_n\le \lim t_n$.

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    In the italian literature, we call these facts "Teoremi della permanenza del segno". The proofs are exactly those you suggest.2012-08-26

2 Answers 2

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Q.2 follows from Q.1, using $a=0$ and the sequence of general term $t_n-s_n$.

For a proof of Q.1 not by contradiction, define $s=\lim\limits_{n\to\infty}s_n$ and consider any $b\lt a$. Then $\varepsilon=a-b$ is such that $\varepsilon\gt0$. Since $s_n\geqslant a$ for every $n$ large enough, $|s_n-b|\geqslant s_n-b\geqslant\varepsilon$ for every $n$ large enough, hence $s\ne b$. Since $s\ne b$ for every $b\lt a$, $s\geqslant a$. (Note that this assumes at the onset that the limit $s$ exists, otherwise the conclusion must be downplayed to $\liminf\limits_{n\to\infty}s_n\geqslant a$.)

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    The assertion that $s_n\geqslant a$ for all but finitely many $n$ means that the set $\{n\mid s_n\lt a\}$ is (empty or) finite.2012-09-25
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For Q.1, let $N = \max\{k : s_k < a\}$. Then $n \ge N$ imples $s_n \ge a$.