We say the terms of a
sequence
(a_n) converge to a number
L — written
\lim_{n \to \infty} a_n = L \qquad\text{or}\qquad a_n \to L
— when the terms get, and stay, arbitrarily close to L. The
informal idea of a limit
is the same as for functions; what is new is that the "approach" happens along the discrete
ticks n = 1, 2, 3, \dots going to infinity. To make
arbitrarily close precise, we use a challenge-and-response game with two players,
\varepsilon and N.
\forall\, \varepsilon > 0 \ \ \exists\, N \in \mathbb{N} \ \ \text{such that}\ \ n \ge N \implies |a_n - L| < \varepsilon.
Read it as a duel: an adversary names a tolerance
\varepsilon > 0 — a tiny band of half-width
\varepsilon around L. You must answer
with a cutoff N such that every term from the
N-th onward lands inside that band. If you can always answer, no
matter how small \varepsilon gets, then a_n \to L.
The centerpiece: proving \tfrac1n \to 0 from the definition
The harmonic sequence a_n = 1/n obviously "wants" to head to
0. Let us earn that claim with the
\varepsilon–N definition, with
L = 0. The whole proof is just answering the adversary's
\varepsilon with a concrete N.
Step 1 — Let the adversary fix \varepsilon. Take
any \varepsilon > 0, however small. Our job is to produce an
N that works for this \varepsilon.
Step 2 — Write down what "inside the band" demands. Since all terms are
positive, |a_n - 0| = \left|\tfrac1n\right| = \tfrac1n, so the
target inequality |a_n - L| < \varepsilon is simply:
\frac{1}{n} < \varepsilon.
Step 3 — Solve it for n. Both sides are positive,
so we may take reciprocals (which flips the inequality):
\frac{1}{n} < \varepsilon \iff n > \frac{1}{\varepsilon}.
Step 4 — Choose N. We need a natural number
bigger than 1/\varepsilon. The
Archimedean property
guarantees one exists; an explicit choice is
N = \left\lfloor \frac{1}{\varepsilon} \right\rfloor + 1, \qquad\text{so}\qquad N > \frac{1}{\varepsilon}.
Step 5 — Verify the cutoff works. Suppose
n \ge N. Then n \ge N > 1/\varepsilon,
and running Steps 2–3 backwards (reciprocate again):
n > \frac{1}{\varepsilon} \implies \frac{1}{n} < \varepsilon \implies \left|\frac1n - 0\right| < \varepsilon.
Step 6 — Collect. For every \varepsilon > 0 we
exhibited an N (namely
\lfloor 1/\varepsilon\rfloor + 1) past which every term lies within
\varepsilon of 0. That is exactly the
definition, so
\lim_{n \to \infty} \frac{1}{n} = 0. \qquad \blacksquare
Concretely: hand me \varepsilon = 0.01 and I answer
N = 101; hand me \varepsilon = 10^{-6}
and I answer N = 1000001. There is always an answer, which is the
whole point.
A sequence (a_n) converges to
L \in \mathbb{R}, written
\lim_{n\to\infty} a_n = L, if
\forall\, \varepsilon > 0 \ \ \exists\, N \in \mathbb{N} : \ \ n \ge N \implies |a_n - L| < \varepsilon.
-
The limit, when it exists, is unique — a sequence cannot converge to two
different numbers.
-
N is allowed to depend on
\varepsilon (smaller tolerance, later cutoff); only
finitely many terms may sit outside the band.
-
A sequence with no such L is said to diverge.
Divergence: 1/n versus (-1)^n
The definition also tells us when a limit fails — compare two sequences side by
side. The point is that
a limit can fail to exist
even when the terms stay perfectly bounded.
Convergent. a_n = 1/n settles into every band, as
we just proved.
Divergent by oscillation. a_n = (-1)^n reads
-1, 1, -1, 1, \dots forever. No single L
can work: pick the band \varepsilon = \tfrac12. To capture the
+1 terms L would need
|1 - L| < \tfrac12, and to capture the
-1 terms it would need
|{-1} - L| < \tfrac12 — but those two intervals are disjoint, so
no L sits in both. Past any cutoff
N there are still +1's and
-1's, so the band is never entered for good. Hence
(-1)^n diverges, though it stays bounded.
You rarely run the \varepsilon–N duel
by hand twice. Once a few limits are known, the
limit laws
carry over verbatim to sequences. If a_n \to A and
b_n \to B, then
a_n + b_n \to A + B, \quad a_n b_n \to AB, \quad \frac{a_n}{b_n} \to \frac{A}{B}\ (B \ne 0).
So, knowing 1/n \to 0, we get
3 + \tfrac{5}{n} \to 3 and
\tfrac{1}{n^2} = \tfrac1n \cdot \tfrac1n \to 0 for free — no new
\varepsilon-juggling.
The other workhorse is the squeeze theorem: if
b_n \le a_n \le c_n for all large n,
and both flanks converge to the same limit
b_n, c_n \to L, then the trapped sequence has no choice but
a_n \to L.
For example 0 \le \tfrac{\sin n}{n} \le \tfrac1n in absolute
value, and both edges go to 0, so
\tfrac{\sin n}{n} \to 0 despite the wobbling numerator.