# Suppose $(s_n)$ converges and that $s_n \geq a$ for all but finitely many terms, show $\lim s_n \geq a$

I've got a few questions about the problem.

Prob :Suppose $(s_n)$ converges and that $s_n \geq a$ for all but finitely many terms, show $\lim s_n \geq a$

The solution here breaks this problem up into two parts.

Q1. I don't understand why is it necessary to consider the finitely many terms that $s_k < a?$

Doesn't the condition that $s_n \geq a$ for all, but finitely many terms and $(s_n)$ being convergent imply that $$\forall \epsilon > 0, \exists N_0 \in \mathbb{N} \implies \forall n > N_0 \implies |s_n - s| <\epsilon$$

So that all the terms after $N_0$ are going to be close to $s$ and therefore $$a < s+\epsilon$$

So why do we need to show the existence of $N > \max \{ M, N_0 \}$ when the definition says there is going to be an $N'$ that gives us convergence?

Q2. Also what is wrong with the following "proof"?

Proof Since $(s_n)$ converges, $$\lim s_n \geq \lim a = a.$$

Ii want to say the proof is wrong because $s_n \geq a$ is not true for every $n$, if it is true for all $n$ then it may be correct? But I thought limits only care about what happens in the long term, so I am not entirely sure what is really the mistake...

EDIT: I was just going to say (very roughly) $s_n \to s \implies \forall \epsilon >0, \exists N_1 \ni \forall n > N_1 s_n < s + \epsilon.$

Now for the rest of the $s_k < 1$ (where $k > N_1$), choose a new $N > \max \{ \max_{k}, N_1 \}$ so that $a \geq s_n < s + \epsilon.$ This is true for all $\epsilon >0$, so take $\epsilon = 0$

Q1. I might be mistaken, but I think this is necessary (at least, for a rigorous proof) because $\forall n > N_0$ $| s_n - s| < \epsilon$ allows $0 < s - s_n < \epsilon \Leftrightarrow s_n < s$; while we know that there exists such $N_0$ that $s_n$ is close to $s$ when $n > N_0$, nothing assures that this $N_0> i$ (for those finitely many $i$ such that $s_i < a$), so why it would be that $\lim_{n\to\infty} s_n \geq a$? At least, I can't see why this would immediately follow. We have to deal with those $s_i$, and show that they don't matter. For one possibly way of doing that which differs a little from the solution you linked to, see for example marty's answer).

EDIT. Okay, I think that works, too (quite elegantly, even). However, I'd argue that it is a substantial step in the proof: the main difference (and point) between this theorem and the more basic one mentioned below ("suppose $s_n > a$ for all $n$", which is usually proved in the textbooks) is that for finite number of elements $s_n < a$, and showing why yet $\lim_{n\to\infty} s_n > a$. In the solution you linked this is dealt with a proof by contradiction, but you already noted the similarities in the argument in the comments.

Someone more proficient than me might be able to provide a more refined explanation why to address this step with a care (or disagree with me), but to me it strikes as a natural thing to do (in other words, the point of proving this theorem as a separate exercise in the first place). (end of edit.)

Q2. Well, I believe you're correct in both of your sub-questions: You can't deduce (in this straightforward fashion) that $\lim_{n\to\infty} s_n \geq \lim_{n\to\infty} a = a$, if for some $i$, $s_i <a$, but if $s_n > a$ for every $n$, then it is straightforward to show that $\lim s_n > \lim a = a$.

Generally speaking, you've got a right idea in that "limits only care about what happens in the long term", but (1) as a statement, it's quite a handwave-y one and (2) that result has to be established before you can use it, which brings us back to Q1.

• I agree that "nothing assures that this $N_0 > i$", but the convergence condition states there will be an $N_0$ that assures $N_0 > i$. – Hawk Sep 14 '14 at 2:04
• "Generally speaking, you've…", okay I can kind of accept this. I feel like going back and forth with this proof.+1 for your input sir. – Hawk Sep 14 '14 at 2:06
• @sidth If it does, then please elaborate on that step of reasoning, which might complete the proof. (As I said, I don't see an immediate step. Rigorous proofs are for simpletons like me.) – kekkonen Sep 14 '14 at 2:09
• Well actually what I wrote there contains 90% of the my argument. The last 10%, I may try to convince you that "I don't need to care about the terms that aren't $\geq a$" because "there may be terms $s_n < a$ when $n > N_0$, but that does not matter much because you can choose a bigger $N$". – Hawk Sep 14 '14 at 2:14
• Huh, I feel like what i just said there is exactly the same as what the solution is saying... – Hawk Sep 14 '14 at 2:14

Suppose $\lim s_n < a$. Let $L = \lim s_n$. Since $L < a$, if $c = a-L$, then $c > 0$.

Since $L = \lim s_n$, by the standard definition of limit, there is an $N$ such that $|L-s_n| < c/2$ for all $n > N$. Rewrite this as if $n > N$, $-c/2 < L-s_n < c/2$ or $s_n < L+c/2 =L+(a-L)/2 =(a+L)/2 < a$ since $L < a$.

Therefore $s_n < a$ for all but a finite number of terms.

Restating, if $\lim s_n < a$, the $s_n < a$ for all but a finite number of terms. By good old contrapositive, if $s_n \ge a$ for all but a finite number of terms then $\lim s_n \ge a$.

• Um this doesn't answer my questions... – Hawk Sep 14 '14 at 1:46