Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $\lim s_n\le \lim t_n$.

share|cite|improve this question
What is contraction? – timur Jul 27 '12 at 4:22
Q.1 follows immediately from the definition of a limit, assuming that $s_n$ has a limit, of course. Q.2 follows immediately from Q.1 by looking at the sequence $t_n-s_n$ and the constant $a=0$. – copper.hat Jul 27 '12 at 4:26
In the italian literature, we call these facts "Teoremi della permanenza del segno". The proofs are exactly those you suggest. – Siminore Aug 26 '12 at 8:37
up vote 0 down vote accepted

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 $\liminf\limits_{n\to\infty}s_n\geqslant a$.)

share|cite|improve this answer
Actually, i got a question, what do finitely many $n$ mean? what about infinitely many $n$, why the condition is needed?? – Mathematics Sep 25 '12 at 13:42
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. – Did Sep 25 '12 at 13:53

For Q.1, let $N = \max\{k : s_k < a\}$. Then $n \ge N$ imples $s_n \ge a$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.