Prove that if $s_n ≤ b$ for all but finitely many $n$, then $\lim s_n ≤ b$.

Question

The question asks me to prove that if $s_n ≤ b$ for all but finitely many $n$, then $\lim s_n ≤ b$ where $(s_n)$ be a sequence that converges. . Here is how I did it but im not sure if its entirely correct. I used proof by contradiction.

Suppose $\lim s_n>b$ and $s_n \leq b$ for all but finitely many $n$. Let $S=\lim s_n$. By definition we have for every $n>N$ which implies $|s_n-s|<\epsilon$. Let $\epsilon= b-2s+s_n$. Now we get $s_n-s<\epsilon=b-2s+s_n$ which simplifies to $s<b=\lim s_n<b$ which is a contradiction.

If the only assumption is that $s_n\le b$ for all but finitely many $n$, we do not know that $\lim s_n$ need exist. Also, do you mean $\lim_{n\rightarrow \infty} s_n$? — vadim123, Jun 04 '13 at 21:01
You should define $\epsilon$ before you define $N$, since $N$ is not independent of $\epsilon$. — Thomas Andrews, Jun 04 '13 at 21:02
So should it go Suppose $lims_n>b$ and $s_n≤b$ for all but finitely many n. Let $s=lims_n$ and let $ϵ=b−2s+s_n$. By definition there exists a natural number $N$ such that $n>N$ and $|s_n−s|<ϵ$. Now we get $s_n−s<ϵ=b−2s+s_n$ which simplifies to $s<b=lims_n<b$ which is a contradiction?. — user60887, Jun 04 '13 at 21:13

score 1 · Answer 1 · answered Jun 04 '13 at 21:18

1

Let $$S=\{n\in\mathbb N\quad|\quad s_n>b\}$$ This set is finite subset of $\mathbb{N}$ so let $n_0=\max S$ hence we have $s_n\leq b\quad \forall n>n_0$ and we conclude.

answered Jun 04 '13 at 21:18

would my proof be correct as well? The one in the comment section. – user60887 Jun 04 '13 at 21:20
You expressed $\epsilon$ in terms of $s_n$ which isn't allowed – Jun 04 '13 at 21:23
well I thought since $s_n$ is a real number it would work. – user60887 Jun 04 '13 at 21:25
You should know that $\epsilon$ in the defintion of limit or continuity or other notions are arbitrary but fixed positive number. – Jun 04 '13 at 21:29
oh yeah thats right. $s_n$ could be negative too. – user60887 Jun 04 '13 at 21:31
How does one conclude from this? I'm unable to follow. – sprajagopal Aug 18 '14 at 08:56

score 1 · Answer 2 · answered Jun 04 '13 at 21:26

Your idea of proof by contradiction might be better modified as follows:

If the limit of $s_n$ is greater than $b$, it can be written as $b+c$ for some $c \in \mathbb{R}^{+}$.

Since this is the limit of the sequence, we can find infinitely many $s_n$ that are arbitrarily close to $b+c$; in particular, we can find infinitely many $s_n$ that are greater than $b$, which contradicts the given that there are only finitely many such $s_n$.

Therefore, our supposition that the limit is greater than $b$ must have been incorrect, hence the limit is less than or equal to $b$ as desired. QED

How do you say that there can be infinitely many $s_n$ near $b+c$? Is there a certain theorem or statement that you can supply for this? — sprajagopal, Aug 18 '14 at 07:02

Prove that if $s_n ≤ b$ for all but finitely many $n$, then $\lim s_n ≤ b$.

2 Answers2