The definition, unpacked
Here is the epsilon–N definition. We say (a_n) converges to L, written a_n -> L, if: for every epsilon > 0 there exists a natural number N such that for all n > N we have |a_n - L| < epsilon. The quantity |a_n - L| is the absolute value of the gap, the distance from a_n to L. So the line |a_n - L| < epsilon reads “the n-th term is within epsilon of L.”
The order of the quantifiers is everything. The for every epsilon comes first; only then does the there exists N answer it. So N is allowed to depend on epsilon — smaller epsilon, larger N — exactly the “starting index moves later” pattern from guide 1. Reverse the order and you would be claiming one N works for all epsilon, which is far too strong and almost never true.
A first proof, written out
The standard pattern: let epsilon > 0 be arbitrary, do scratch work to discover how large n must be, then write the proof forward by choosing N. Here is the full proof that 1/n -> 0. Notice the structure — the scratch work is where you find N, the clean proof is where you justify it.
Claim: 1/n -> 0. Scratch (find N): we want |1/n - 0| < e. |1/n - 0| = 1/n (since n > 0). 1/n < e <=> n > 1/e. So any N with N >= 1/e will do. Proof: Let e > 0 be given. By the Archimedean property, choose a natural N with N > 1/e. Let n > N. Then n > 1/e, so 1/n < e. Hence |1/n - 0| = 1/n < e. Since e > 0 was arbitrary, 1/n -> 0. QED
Uniqueness, and proving divergence
The definition immediately earns its keep. A convergent sequence has only one limit. Suppose a_n -> L and a_n -> M with L not equal to M. Set epsilon = |L - M| / 2 > 0. From some point on, a_n is within epsilon of L; from some point on, within epsilon of M. Past both points the triangle inequality gives |L - M| <= |L - a_n| + |a_n - M| < epsilon + epsilon = |L - M|, so |L - M| < |L - M|, a contradiction. Hence L = M and the limit is well-defined.
To prove divergence you negate the definition. The sequence does NOT converge to L means: there exists some epsilon > 0 such that for every N there is an n > N with |a_n - L| >= epsilon. One bad epsilon, infinitely often, is enough. Watch this kill b_n = (-1)^n.
Claim: a_n = (-1)^n does NOT converge.
Suppose toward contradiction a_n -> L for some L.
Take e = 1 in the definition. Then for some N,
|(-1)^n - L| < 1 for all n > N.
Pick an even n > N and an odd m > N (both exist). Then
|1 - L| < 1 and |-1 - L| < 1.
By the triangle inequality,
2 = |1 - (-1)| = |(1 - L) - (-1 - L)|
<= |1 - L| + |-1 - L| < 1 + 1 = 2.
So 2 < 2, a contradiction.
Therefore no such L exists: (-1)^n diverges. QED