A sequence that settles cannot run away, and the proof has a shape worth learning because it is used again and again: the definition controls the tail, a finite list is handled by taking a maximum, and the two are put together. The converse is false, and the sequence that shows it is the one from the last lesson — bounded, and settling nowhere.

A ship that has entered harbour for good is somewhere on the chart. The definition says it is inside the harbour from some day onwards, which handles all but finitely many days; the days before that are a finite list, and a finite list has a farthest point. The chart is then drawn to cover both.

From $N$ on, the terms are within one of the limit, so their size is at most $\lvert L\rvert+1$. Before $N$ there are finitely many terms, and a finite list has a largest size. The bound $M$ is the larger of the two.

If $a_n\to L$ then there is an $M$ with $\lvert a_n\rvert\le M$ for every . The proof applies the definition once, at the convenient tolerance , and then handles the finitely many early terms separately. The converse fails, so boundedness is a necessary condition for convergence and not a sufficient one — which makes its contrapositive the quickest test for divergence there is.

• The contrapositive is the working form: an unbounded sequence converges to nothing. $(n)$, $(n^2)$ and $((-1)^n n)$ all diverge for this reason alone.

Use the definition once, at a tolerance chosen for convenience rather than because anything demands it. With the tolerance equal to one, every term past some place lies within one of the limit, so the size of each is at most the size of the limit plus one. That leaves only the terms before that place, and there are finitely many of them, so their sizes have a largest. The bound is whichever of the two numbers is larger.

Any fixed tolerance would do; one is convenient.

The definition controls every term from some place onwards.

The triangle inequality turns closeness to L into a bound on the size.

The remaining terms are a finite list, so their sizes have a largest.

Whichever of the two is larger bounds both parts at once.

When the definition is being used rather than verified, the tolerance is yours to pick. Pick one that makes the arithmetic easy.

Changing finitely many terms changes no limit, and changing them cannot break boundedness either.

• A convergent sequence is bounded, and the proof controls the tail and then the finite front.
• The converse is false, and the alternating sequence shows it.
• The contrapositive is the quickest test for divergence.
• Both convergence and boundedness survive a change to finitely many terms.

Great work! You now know how to:

• Prove that a convergent sequence is bounded
• Use the contrapositive as a test for divergence
• Say why the finite front must be separated from the tail
• Give a bounded sequence with no limit