How to properly construct an $\epsilon-N$ proof I've asked a couple of questions now on this type of proof.
My question is: could someone give me a step-by-step set of steps to follow in the general case (e.g. for proof by induction, I'd say the 3 main steps are: check the statement's true in the base cas, assume the result is true for $n$, and then finally show $P(n)\implies P(n+1)$).
Is there a general recipe for $\epsilon-N$ proofs?
All I know is that we need to show that $|a_n-L|<\epsilon$.
 A: The general structure of a proof by induction is as follows. (Here $P$ is some propery of natural numbers.) Note that there are 2 (not 3) essential steps - assuming $P(n)$ and using some reasoning to get $P(n+1)$ is part of the same step.
Theorem. For all $n \in \mathbb N$, $P(n)$.
Proof.


*

*$\dots$ some reasoning $\dots$. Therefore $P(0)$.

*Take $n \in {\mathbb N}$ and assume that $P(n)$. $\dots$ some reasoning $\dots$ Therefore $P(n+1)$.


Hence, by induction, $P(n)$ holds for all $n \in \mathbb N$.

The general structure of an $\epsilon$-$N$ proof is as follows. (Here $p(n)$ is some expression in $n$ that should eventually become small).
Theorem. For all $\epsilon > 0$, there is an $N \in \mathbb N$ such that for all $n \geq N$, $p(n) < \epsilon$.
Proof. Let $\epsilon > 0$. Take $N = \dots$ some clever expression in $\epsilon$ $\dots$ and take $n \geq N$.
$\dots$ some reasoning. Therefore $p(n) < \epsilon$.
In practice, a tricky part is to figure out what some clever expression in $\epsilon$ should be. It often is beneficial to leave that open and only at the very end out figure what you should have taken there to make $p(n) < \epsilon$ hold.
A: Regarding your question about $\epsilon-N$ proofs there is a general framework, but not a general recipe that you could follow that will always allow you to figure a good upper bound.
The proper definition of a convergent sequence is this: $\{a_n\}_{n\geq 1}$ is a sequence that converges to $L$ iff for every $\epsilon>0$ there exists a $N(\epsilon)\in\mathbb{N}$ such that if $n\geq N(\epsilon)$ then $|a_n-L|<\epsilon$. In real analysis you will encounter a lot of definitions with $\epsilon$ and it is always beneficial to really try to understand what is going on. This is the very first definition that most students come across dealing with an epsilon. What is really going on in this definition is that if I fix any $\epsilon>0$ then by the Archimedean property there exists a natural number $N$ that depends on epsilon (Think of $N$ as a position marker) such that for all $n$ bigger then $N$ then $|a_n-L|<\epsilon$. This last part just means that the distance between $a_n$ and $L$ is getting closer and closer while being less than $\epsilon$. 
The best thing to do is work through a lot of examples and problems on your own. Some of the more harder ones require the use of the binomial theorem or Bernoulli's inequality if you are just going straight off this definition. Here is one example of a problem and I will show you the way I approach them. 
Suppose I want to show that $\lim_{n\to\infty}\frac{6n}{n^2+1}=0$ where $n$ is a natural number.
First step I do some scratch work. The key to any of these problems is finding a suitable upper bound. So working backwards in the definition I see that $|\frac{6n}{n^2+1}-0|=|\frac{6n}{n^2+1}|=\frac{6n}{n^2+1}\leq \frac{6n}{n^2}=\frac{6}{n}<\epsilon$. Now I want to isolate $n$ by itself so $\frac{6}{n}<\epsilon\implies n>\frac{6}{\epsilon}$. (So now taking my work in reverse I can come up with a proof).
Proof:
Let $\epsilon>0$( This is where I fix epsilon). It follows that $\frac{1}{\epsilon}>0\implies \frac{6}{\epsilon}>0$. Then by the Archimedean property there exists a 
$N\in\mathbb{N}$ such that $N>\frac{6}{\epsilon}$. Then if $n\geq N>\frac{6}{\epsilon}$ then $|\frac{6n}{n^2+1}-0|\leq \frac{6}{n}< 6(\frac{\epsilon}{6})=\epsilon$ (The last follows from the fact that since  $n\geq N>\frac{6}{\epsilon}\implies n>\frac{6}{\epsilon}\implies\frac{1}{n}<\frac{\epsilon}{6}\implies \frac{6}{n}<\epsilon$). Now this about as detailed as I can get and once you get better at these your proofs will be much shorter. Now this was a simpler problem. I'd suggest taking look at harder ones that use the binomial theorem or Bernoulli's inequality in order to find an upper bound as these are very useful techniques. 
A: The whole 'philosophy' of this is the limit. What you have to do is to see does $|a_n-L|$ tends to $0$ as n tends to infinity. Than replace that by the definition of the limit. If else $a_n$ does not tend to $L$.
