To show a sequence does not converge to a number, deny the definition. Each quantifier turns over: instead of every tolerance being answerable, one tolerance is not; instead of a place beyond which the terms behave, every place is followed by a term that misbehaves. Naming that one stubborn tolerance is the whole proof.

Back to the challenge. To win as the challenger you do not have to defeat every answer; you have to name a single margin your opponent can never meet, and then, wherever they point in the list, produce a later term outside it.

This is the negation of a statement with three nested quantifiers, done exactly as the chapter on logic set out. Each quantifier flips as the negation moves inward, and the inequality reverses at the end.

• One tolerance is enough. You choose it, and it may be as convenient as you like.
• It must defeat every place, so the argument has to work for an arbitrary $N$.
• For that $N$ you produce one later term outside the tolerance, not all of them.
• A sequence that converges to no number at all is called divergent.
• $a_n\to\infty$ means: for every $$ there is an with for all . Such a sequence diverges — the notation records how, and does not name a limit.

Here $a_n = (-1)^n$ and $L = 1$, the value it keeps returning to. Take : wherever you point in the list, a term at comes later, a distance 2 away. One tolerance with no answer is the whole denial — and no other choice of escapes it, since some term is always at least 1 away.

Suppose a limit existed. With a tolerance of one, terms of both signs would have to lie within one of it, and the triangle inequality then squeezes the distance between those two terms below its own value.

Assume a limit exists and drive it to an impossibility.

One tolerance is all the denial needs, and this one is convenient.

This is what convergence would give for that tolerance.

Even places give one, odd places give minus one, and both kinds occur forever.

The triangle inequality routes the distance between the two terms through $L$.

The supposition produced a contradiction and must be rejected.

Three flips, then the condition inside is denied.

For $((-1)^n)$, taking $\varepsilon = 1$ is enough.

Terms one apart cannot both lie within one of the same number.

The place was handed over, and a misbehaving term was produced beyond it.

$((-1)^n)$ is bounded and divergent.

It never leaves the range from minus one to one, and it never settles either.

A sequence can fail to converge in an orderly way. The notation says how it fails, not that it has a limit.

• To deny convergence, produce one tolerance that defeats every place.
• The place is your opponent’s: the argument must work for an arbitrary N.
• For that place you produce one later escaping term, not all of them.
• Divergence to infinity is a way of failing, not a limit.

Great work! You now know how to:

• Negate the definition of a limit and use the result
• Choose one convenient tolerance and defeat an arbitrary place
• Tell divergence from unboundedness, and read the arrow to infinity correctly