The Monotone Convergence Theorem
Why it must be true. Picture an increasing sequence that never passes a
ceiling M. Each term is at least as big as the last, yet none can
climb above M. Crammed into the finite room between
a_1 and M, with no backtracking
allowed, the terms have nowhere to go but to pile up against a single value at the
top. That value is the limit.
The deep reason is completeness. "Pile up against a single value at the top"
is precisely the statement that the set of terms has a least upper bound. The reals
are built to guarantee one exists: every non-empty set bounded above has a
supremum in \mathbb{R}. That axiom —
completeness
— is the whole engine. Here is the argument in steps.
Step 1 — Collect the terms and take the supremum. Let
(a_n) be increasing and bounded above. The set of its values
S = \{a_n : n \ge 1\} is non-empty and bounded above by
M, so by completeness it has a least upper bound:
L = \sup S = \sup_{n} a_n.
Step 2 — Fix a tolerance. Let \varepsilon > 0 be
given. We will produce a cutoff N past which every term is within
\varepsilon of L.
Step 3 — Use the "least" in least upper bound. Because
L is the smallest upper bound, the slightly smaller number
L - \varepsilon is not an upper bound. So some term must
poke above it: there is an index N with
a_N > L - \varepsilon.
Step 4 — Let monotonicity carry it forward. The sequence is increasing, so
for every n \ge N we have a_n \ge a_N > L - \varepsilon.
And L is an upper bound, so a_n \le L.
Squeezing both sides:
L - \varepsilon < a_n \le L < L + \varepsilon \qquad (n \ge N).
Step 5 — Read off convergence. That sandwich says
|a_n - L| < \varepsilon for all
n \ge N. Since \varepsilon was
arbitrary, the ε–N definition is satisfied:
\lim_{n \to \infty} a_n = \sup_{n} a_n. \qquad \blacksquare
A decreasing sequence bounded below converges to its infimum by the mirror argument.
Let (a_n) be a monotone sequence of real numbers.
-
If (a_n) is increasing and bounded above, it
converges, and \displaystyle\lim_{n\to\infty} a_n = \sup_n a_n.
-
If (a_n) is decreasing and bounded below, it
converges, and \displaystyle\lim_{n\to\infty} a_n = \inf_n a_n.
-
A monotone sequence that is unbounded diverges to
\pm\infty; so for a monotone sequence,
bounded ⇔ convergent.
-
The existence of the limit rests entirely on the
completeness of \mathbb{R} — the least-upper-bound
axiom. Over \mathbb{Q} the theorem is false.
Worked example: a_n = \left(1 + \tfrac1n\right)^n climbs to e
The most famous application defines e itself. The sequence
a_n = (1 + 1/n)^n starts at
a_1 = 2, then
a_2 = 2.25,
a_3 \approx 2.370,
a_4 \approx 2.441, … We show it satisfies both hypotheses.
Increasing. Expanding by the binomial theorem,
a_n = \sum_{k=0}^{n} \binom{n}{k}\frac{1}{n^k} = \sum_{k=0}^{n} \frac{1}{k!}\prod_{j=0}^{k-1}\!\left(1 - \frac{j}{n}\right).
Going from n to n+1, each factor
(1 - j/n) grows (closer to 1), and the
sum gains an extra positive term — so a_{n+1} > a_n.
Bounded above. Each factor in the product is at most
1, so a_n \le \sum_{k=0}^n \tfrac{1}{k!},
and using k! \ge 2^{k-1},
a_n \le 1 + \sum_{k=1}^{n} \frac{1}{2^{k-1}} < 1 + 2 = 3.
Increasing and bounded above by 3 — the Monotone Convergence
Theorem fires, and the limit exists. We christen it e:
e = \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n \approx 2.71828.
Monotone Convergence shines on
recursively defined
sequences, where no closed form is in sight. Consider
a_1 = \sqrt{2}, \qquad a_{n+1} = \sqrt{2 + a_n}.
Bounded above by 2: if a_n < 2 then
a_{n+1} = \sqrt{2 + a_n} < \sqrt{4} = 2, so by induction every
term stays under 2. Increasing: one checks
a_{n+1} > a_n \iff 2 + a_n > a_n^2 \iff (a_n - 2)(a_n + 1) < 0,
true whenever 0 < a_n < 2. Increasing and bounded above ⇒ the
limit L exists.
Now find it. Taking n \to \infty in the recurrence (both sides
converge to the same L), L must be a
fixed point:
L = \sqrt{2 + L} \implies L^2 - L - 2 = 0 \implies (L-2)(L+1) = 0,
and since the terms are positive, L = 2. The theorem supplied
existence; the algebra of the fixed point pinned the value.