Proving a sequence converges $\epsilon-N$

It seems my biggest nightmare has come to haunt me again! I first came across the formal proof of a limit $\epsilon-\delta$ in Calculus 1, I never truly mastered it since at the time it was just racking my brain. I have now begun self-teaching a course in sequences & series and it has already come up. It has also been emphasised to me that being able to understand and do this is fundamental for success in this course.

I'm almost 100% certain that I'd have no trouble applying the methods, I've seen in tutorials to prove a sequence converges. However, I don't understand how the $\epsilon-N$ proof actually proves a sequence converges; if that makes much sense.

Here are a few questions, I'd like to ask:

• What does this statement mean: $|a_n - L| < \epsilon$ L refers to the limit.
• Why does this proof actually prove a sequence converges?

• Why does $\epsilon>0$?

Those are the only questions that currently come to mind, also if anyone can provide any additional advice on how I can wrap my head around these proofs please feel free to post!

• For a sequence's convergence the definition is $\varepsilon$ - $N$ rather than $\varepsilon$ - $\delta$. Check your textbook. – hardmath Dec 12 '16 at 1:51
• Geometrically, what does the value of $|b-a|$ tell you if a and b are real numbers? – user328442 Dec 12 '16 at 1:54
• @user328442 from my limited knowledge on the absolute value function, I think it tells me the distance $b-a$ is from 0 – idk Dec 12 '16 at 1:58
• That's true although another way to look at it is that it's the distance between a and b. – user328442 Dec 12 '16 at 1:59
• Apply this idea to $|a_n - L| < \epsilon$. For $\epsilon > 0$, what does the statement mean, geometrically? – user328442 Dec 12 '16 at 2:00

The definition is the following $$\forall \epsilon >0, \exists n_0\in\mathbb{N} : |a_n-L|\leq \epsilon \,\,\,\forall n\in\mathbb{N}, n>n_0$$

$|a_n-L|$ represents the distance between the value of the limit to which the sequence converges and the $n^{nt}$ element of the sequence. It is a distance in absolute value because $a_n$ could be bigger or smaller than $L$ (depending if the sequence is decreasing or increasing or alternating), so we take the Absolute value to always have a positive value.

$|a_n -L|<\epsilon$ means that this distance is smaller than any value of $\epsilon >0$ no matter how small you take it to be! Recall that if you are doing analysis there should be a lemma saying that if $|x|<\epsilon$, $\forall \epsilon > 0$ $\implies |x| = 0$. Therefore $|a_n -L|<\epsilon$ means that you want this distance to become zero!

The rest of the definition it is saying that if a sequence converges to $L$, i.e. if the $n^{nt}$ term, as $n\to\infty$ goes to $L$ , then this means that after a certain point ($n>n_0$, so $n_0$ is the "turning point") all the terms of the sequence coming after $a_{n_0}$ can approach $L$ so much that the distance between $L$ and the term you've chose, after $a_{n_0}$ is smaller than the given epsilon.

What you do in the proving it, is exactly this: you choose an arbitrary $\epsilon >0$. Then what happens is that you can always find a $n_0$ (which is the subscript of an element of the sequence, say $n_0 = 18923842$, this means that $a_{18923842}$ is the element you're talking about) such that every term after it (i.e. $a_{18923843}$ for instance) has a distance to $L$ which is a value (positive because of the absolute value) smaller than the epsilon you chose before!

• I'm still reading it, trying to make complete sense of it all. First two paragraphs have registered and now make complete sense. So, I'd just like to say thank you! I think you saved me from getting grey hairs at 16!!! – idk Dec 12 '16 at 2:37
• @idk don't worry, there's a typo, $n^{th}$ clearly! Tell me what I didn't explain clearly and I'll try to explain it better! – Euler_Salter Dec 12 '16 at 2:40
• It is basically like a game, player 1 chooses $\epsilon >0$ randomly. Player 2 knows the sequence converges, so picks a natural number $n_1$. Player 1, takes this number and checks what element of the sequence corresponds to $n_1$, i.e. $a_{n_1}$ and tells this to player 2. Player 2 then checks if for every element of the sequence that comes after $a_{n_1}$, $|a_n- L|<\epsilon$, where $n>n_1$. If it is true, then you're done. If this is false, player 2 picks a bigger natural number, call it $n_2$ (while Player 1 keeps the same $\epsilon$) and the game goes on until you find $n_0$. – Euler_Salter Dec 12 '16 at 2:47
• confused again lol. I'm very dopey sorry, but I don't understand why you have to check that every element after $a_n1$ satisfies $|a_n - L|<\epsilon$ – idk Dec 12 '16 at 3:11
• Is not that you have to. This is a definition, people made it. It is written this way because the message that it conveys is: If a sequence converges, then if I go far enough in the sequence, I'll get to a point where all the elements have a distance from $L$ smaller than a given $\epsilon$. Which means that if you look at elements far enough, they will be indistiguishible from the actual limit, and that agrees with out intuition! – Euler_Salter Dec 12 '16 at 3:15