Imagine climbing a staircase inside a room. Every step takes you upward — or at worst leaves you level — and yet no step can ever carry you above the ceiling. You might not know where you will come to rest, but come to rest you must: forever rising with nowhere left to rise to, you are squeezed against some final height. That squeeze is a theorem — the Monotone Convergence Theorem — and it is the workhorse existence theorem of analysis: it proves a limit exists without you ever having to name it.

Why is that such a big deal? Because running the ε–N duel requires you to already know the limit L you are aiming at. Often we do not — we only know the shape of the sequence. Two structural features, both checkable by bare-hands algebra, are enough to guarantee a limit is waiting at the end.

A sequence (a_n) is monotone if it only ever moves one way:

It is bounded above if some ceiling M is never exceeded, a_n \le M for all n (and bounded below symmetrically, with a floor). The headline result is that monotone plus bounded together force convergence — the staircase must stop somewhere under the ceiling.

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:

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

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:

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:

A decreasing sequence bounded below converges to its infimum by the mirror argument.

Watch it climb to the ceiling

Here is an increasing sequence a_n = L\,(1 - c^{\,n}) creeping up toward a ceiling — the dashed line is its supremum L = 2, the least upper bound the terms never cross but get arbitrarily close to. Turn the climb rate up and the terms reach the ceiling sooner; turn it down and they crawl. Either way, bounded plus monotone means a limit waits at the top.

Notice something subtle: 3, 7 and a million are also upper bounds for these dots, and any one of them is good enough to fire the theorem. The theorem does not care how sloppy your ceiling is — but the limit itself is always the least upper bound, the one the dots actually kiss. Proving convergence with a crude bound and then pinning down the exact limit separately is not cheating; it is the standard workflow, as the examples below show.

Worked example 1 — certifying "monotone" by hand

Before the theorem can fire, you must prove monotonicity, and there are two standard moves. The difference test: show a_{n+1} - a_n \ge 0 (increasing) or \le 0 (decreasing). The ratio test (for positive terms only): show a_{n+1}/a_n \ge 1 or \le 1. Use whichever makes the algebra collapse — differences suit sums and fractions, ratios suit powers and factorials.

By differences: take a_n = \dfrac{n}{n+1}, the sequence \tfrac12, \tfrac23, \tfrac34, \dots Then

so the sequence is (strictly) increasing. And a_n = \tfrac{n}{n+1} < 1 always, so it is bounded above by 1. Increasing + bounded above ⇒ it converges — and here we can even see the limit is 1, the supremum.

By ratios: take a_n = \dfrac{2^n}{n!}, the sequence 2, 2, \tfrac43, \tfrac23, \tfrac{4}{15}, \dots All terms are positive, and

so from the second term on the sequence is decreasing — and it is bounded below by 0, since every term is positive. Decreasing + bounded below ⇒ it converges. (In fact the ratio \tfrac{2}{n+1} \to 0 crushes the terms to the floor: the limit is 0, the infimum.) Note the useful freebie: eventually monotone is enough. The first few unruly terms never affect whether a limit exists, nor what it is.

Worked example 2 — a recursive sequence and its fixed point

Monotone Convergence shines brightest on recursively defined sequences, where each term is cooked from the previous one and no closed formula is in sight. Consider the nested square roots

Step 1 — bounded above by 2, by induction. Base case: a_1 = \sqrt2 < 2. Inductive step: if a_n < 2, then a_{n+1} = \sqrt{2 + a_n} < \sqrt{2+2} = 2. So every term stays strictly under 2: the recursion can never manufacture a term that breaks the ceiling, because the ceiling reproduces itself.

Step 2 — increasing. For positive terms,

which holds precisely when 0 < a_n < 2 — and Step 1 guarantees that. So the sequence climbs.

Step 3 — conclude, then hunt the value. Increasing and bounded above: the Monotone Convergence Theorem fires, and a limit L exists. Only now are we allowed to take n \to \infty on both sides of the recurrence (both sides converge to the same L, and \sqrt{\;\cdot\;} is continuous), which says L is a fixed point:

and since all terms are positive, L = 2. This two-step dance — the theorem supplies existence, the fixed-point equation pins the value — is the standard playbook for recursive sequences, from these nested radicals to Heron's square-root algorithm a_{n+1} = \tfrac12\!\left(a_n + \tfrac{c}{a_n}\right), a decreasing-and-bounded-below sequence whose fixed point is \sqrt{c}. Here the ceiling 2 happened to equal the limit, but that was luck: bounding by 100 would have proved convergence just as well.

Worked example 3 — a_n = \left(1 + \tfrac1n\right)^n climbs to e

The most famous application defines e itself. Compound interest at 100\%, chopped into n ever-smaller instalments, grows a unit deposit to a_n = (1 + 1/n)^n: the sequence starts at a_1 = 2, then a_2 = 2.25, a_3 \approx 2.370, a_4 \approx 2.441, … Does chopping finer and finer make the payout explode? We check the two hypotheses (a sketch — the full details belong to a later page).

Increasing. Expanding by the binomial theorem,

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},

Increasing and bounded above by 3 — the Monotone Convergence Theorem fires, and the limit exists. No compounding scheme ever pays more than triple. The limit has no nicer closed form, so we simply christen it e:

Read that again: one of the most important constants in mathematics is defined as the limit of a monotone bounded sequence. Without this theorem, the symbol e would name nothing at all.