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A bounded monotone sequence converges to its supremum or infimum, so a limit can be produced before its value is known.
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By the end of this lesson you will be able to prove the monotone convergence theorem from completeness, and to use it on a recursive sequence in the right order: guess the fixed point, prove a bound and monotonicity by induction, conclude that a limit exists, and only then take limits in the recursion to identify it. You will also be able to say what goes wrong when those last two steps are swapped.
The definition of a limit, which needs the limit in order to state the condition, and the approximation property of a supremum. Every proof so far has needed the limit named in advance. This lesson is the first that produces one from nothing but boundedness.
A sequence is increasing when each term is at least the one before it, decreasing when each is at most, and monotone when it is one or the other. A recursive sequence is defined by a rule carrying each term to the next, and a fixed point of that rule is a value the rule leaves unchanged.
Monotone convergence. An increasing sequence that is bounded above converges, and its limit is $\sup\{a_n\}$. A decreasing sequence bounded below converges to $\inf\{a_n\}$.
Proof (increasing case). The terms form a non-empty set bounded above, so $L = \sup\{a_n\}$ exists — that is completeness, used once. Let $\varepsilon > 0$. Since $L - \varepsilon$ is below the least upper bound it is not an upper bound, so some $a_N > L - \varepsilon$. By monotonicity $a_n \ge a_N$ for $n \ge N$, and $a_n \le L$ always. So $L - \varepsilon < a_n \le L$, that is $|a_n - L| < \varepsilon$, for every $n \ge N$.
Both hypotheses do distinct jobs. Boundedness produces the candidate; monotonicity delivers the sequence to it. Drop the first and $a_n = n$ increases to nothing; drop the second and $(-1)^n$ is bounded and settles nowhere.
Completeness is what lets a limit be produced rather than guessed. Boundedness plus monotonicity, or the Cauchy condition, names a number before anyone knows what it is; over the rationals the same argument runs and the number it names is not there.
Why this is the practical theorem. The definition cannot be used until the limit is known. Monotone convergence reverses that: it establishes existence from two facts that can be checked on the terms themselves, and the value is identified afterwards.
Another way: picture
Dots climbing to the right, with a ceiling drawn above them. They can never cross the ceiling and never come back down, so they have to pile up against some height — and the lowest ceiling that works is the height they pile up against.
Another way: steps
For a recursive sequence $a_{n+1} = f(a_n)$:
Steps 4 and 5 cannot be swapped.
Let $a_1 = 1$ and $a_{n+1} = \sqrt{2 + a_n}$.
Guess. $L = \sqrt{2 + L}$ gives $L^2 - L - 2 = 0$, so $L = 2$ (the negative root is impossible for a positive sequence).
Bound. $a_1 = 1 < 2$; and if $a_n < 2$ then $a_{n+1} = \sqrt{2 + a_n} < \sqrt{4} = 2$. So $a_n < 2$ for all $n$, by induction.
Monotone. $a_{n+1}^2 - a_n^2 = 2 + a_n - a_n^2 = (2 - a_n)(1 + a_n) > 0$ using the bound. Both terms are positive, so $a_{n+1} > a_n$.
Exists. Increasing and bounded above, so a limit $L$ exists.
Identify. $a_{n+1} \to L$ as well, since it is the same sequence shifted, so $L = \sqrt{2 + L}$ and $L = 2$.
The guess at the start and the identification at the end use the same equation, and only the second one is a step in the proof. Writing 'the limit satisfies $L = \sqrt{2+L}$, so it is $2$' without steps 2 to 4 proves nothing: the same sentence can be written about $a_{n+1} = 2a_n$, whose fixed point is $0$ and which diverges from every non-zero start.
| Monotone | Not monotone | |
|---|---|---|
| Bounded | converges, to the supremum or infimum | may or may not converge |
| Unbounded | diverges to infinity | anything |
Two readings of the table are worth making explicit.
The condition is sufficient, not necessary. $(-1)^n/n$ converges and is not monotone. A sequence failing the test has not been shown to diverge.
A monotone sequence always has a limit in the extended sense. It converges, or it runs off to infinity; it cannot oscillate. That dichotomy is what makes monotone subsequences so useful, and it is why the next lesson starts by extracting one.
Solving the fixed point and stopping. That identifies what the limit would be. It does not show there is one, and for a divergent recursion it hands back a number with no meaning.
Taking the bound to be the limit. Bounded above by $7$ does not mean converging to $7$. The limit is the least upper bound of the terms, which is usually smaller than the bound you happened to prove.
Proving monotonicity without the bound. For most recursions the comparison $a_{n+1} > a_n$ needs the bound from step 2. The induction usually has to carry both facts at once.
Assuming the limit is attained. An increasing sequence never reaches its supremum unless it is eventually constant. The limit is approached, not achieved.
Let $a_1 = 0$ and $a_{n+1} = (a_n + 6)/2$. The fixed point equation $L = (L+6)/2$ gives $L = 6$.
A guess, to be bounded by.
If $a_n < 6$ then $a_{n+1} = (a_n + 6)/2 < 6$, and $a_{n+1} - a_n = (6 - a_n)/2 > 0$.
Bounded above and increasing, both by induction.
Monotone convergence gives a limit; taking limits in the recursion identifies it as $6$.
Only now is the fixed point a theorem.
Let $a_n = (1 + 1/n)^n$. Expanding by the binomial theorem shows the terms increase with $n$.
Monotone, by a computation on the terms.
The same expansion is bounded term by term by $\sum 1/k!$, which is below $3$.
Bounded above, by comparison.
So the sequence converges. Its limit is the number $e$ — and this theorem is what says there is a number there to name.
The definition of $e$ rests on this theorem.
The fixed point equation $L = \frac{1}{2}(L + 2/L)$ gives $L^2 = 2$, so the guess is $\sqrt{2}$.
Guess first, to know what to bound by.
Every term is at least $\sqrt{2}$, because $\frac{1}{2}(x + 2/x) - \sqrt{2} = \frac{(x - \sqrt{2})^2}{2x} \ge 0$ for positive $x$.
Bounded below, by an algebraic identity.
And $a_n - a_{n+1} = \frac{a_n^2 - 2}{2a_n} \ge 0$ using that bound, so the sequence decreases. Monotone convergence then gives a limit, and the recursion identifies it as $\sqrt{2}$ — which is Newton's method for a square root, and a second proof that the number exists.
Build the proof that an increasing sequence bounded above converges to the supremum of its terms.
This task has no paper form; do it on a device.
Let $a_1 = 0$ and $a_{n+1} = \dfrac{a_n + 6}{2}$. The sequence increases and is bounded above by $6$. What is its limit?
Answer:
Put the steps of the standard treatment of a recursive sequence, such as $a_{n+1} = \dfrac{a_n + 3}{2}$, into order.
Number the steps in order (write the number in the box):
Match each sequence to the description that fits it.
| increasing and bounded above | decreasing and bounded below | monotone but unbounded | convergent but not monotone | |
|---|---|---|---|---|
| $a_n = 5 - 1/n$ | ||||
| $a_n = 5 + 1/n$ | ||||
| $a_n = 5n$ | ||||
| $a_n = (-1)^n/n$ |
A sequence increases and every term is below $5$. Which fact is what actually produces its limit?
For which starting values $a_1$ does the sequence $a_{n+1} = \dfrac{a_n + 8}{2}$ increase at its first step?
This task has no paper form; do it on a device.
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Build the proof that an increasing sequence bounded above converges to the supremum of its terms.
This task has no paper form; do it on a device.
You can prove monotone convergence from the completeness axiom and use it to settle a recursive sequence. Say in your own words why solving the fixed point equation on its own proves nothing. Next: what can still be said when a bounded sequence refuses to be monotone.
10. Your turn: $a_1 = 2$, $a_{n+1} = \dfrac{1}{2}\left(a_n + \dfrac{2}{a_n}\right)$, step 3