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A sequence is a function on the whole numbers; a bounded monotone one converges, and the theorem names no limit.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to find the limit of an explicitly given sequence by comparing growth rates or dividing by the dominant term, borrow a limit from the matching function of a real variable when a rule such as L'Hopital's is wanted, state the monotone convergence theorem and give counterexamples showing that neither hypothesis can be dropped, prove a recursive sequence monotone and bounded by induction, and solve the fixed-point equation for its limit only after convergence has been established.
Calculus I defined $\lim_{x\to\infty}f(x) = L$: for every $\varepsilon > 0$ there is an $M$ beyond which $|f(x) - L| < \varepsilon$. A sequence is a function whose inputs are only the whole numbers, and its limit is the same definition with $N$ in place of $M$.
So one useful fact comes free: if $f(x) \to L$ as $x \to \infty$ and $a_n = f(n)$, then $a_n \to L$. That is what permits using L'Hopital's rule on $\dfrac{\ln n}{n}$ — the rule is about functions of a real variable, and the sequence borrows the answer. The converse fails: $\sin(\pi n)$ is always $0$ while $\sin(\pi x)$ has no limit.
A sequence is increasing if $a_{n+1} \ge a_n$ for every $n$, decreasing if $a_{n+1} \le a_n$, and monotone if it is one or the other. It is bounded above if some number exceeds every term, and bounded if it is bounded above and below.
A tail is what is left after discarding finitely many terms. Convergence is a property of tails only: changing the first thousand terms changes nothing.
A sequence is given explicitly by a formula for $a_n$, or recursively by a first term and a rule for getting the next from the last. A recursion may converge without any formula for its terms existing, which is exactly the situation the monotone convergence theorem was made for.
Definition. $a_n \to L$ means: for every $\varepsilon > 0$ there is an $N$ with $|a_n - L| < \varepsilon$ for all $n \ge N$. The sequence converges if some such $L$ exists, and the limit is then unique.
The limit laws carry over unchanged: sums, products, quotients (non-zero denominators) and the squeeze theorem all hold for sequences, with the same proofs.
Growth rates, in increasing order of speed: $$\ln n \ll n^p \ (p>0) \ll b^n \ (b>1) \ll n! \ll n^n.$$ Each loses to the next, so a ratio of two of them tends to $0$ or $\infty$ according to which is on top. This one line answers most limit questions in the chapter.
Monotone convergence theorem. An increasing sequence bounded above converges; a decreasing sequence bounded below converges.
This is the theorem the rest of the unit runs on, and what is remarkable about it is what it does not require: no formula for the terms, and no candidate for the limit. It converts two checkable properties into existence. It is equivalent to the completeness of the real numbers — over the rationals it is false, since $1, 1.4, 1.41, 1.414,\ldots$ is increasing, bounded and has no rational limit.
Recursions. If $a_{n+1} = f(a_n)$ converges to $L$ and $f$ is continuous, then $L = f(L)$. Solving that gives the only possible limit; it does not prove one exists, and the order matters.
Another way: picture
An increasing sequence plotted against $n$, with a horizontal ceiling drawn above it. The terms climb and can never cross the ceiling, so they have to pile up somewhere below it — they cannot keep rising and they cannot come back down. The theorem says the level they pile up at is the lowest ceiling that works, and the picture makes it obvious why no formula is needed to know it is there.
Another way: steps
The monotone convergence theorem asks for monotone and bounded, and it is worth seeing that both are needed, because a theorem with a redundant hypothesis is being misremembered.
Monotone without bounded. $a_n = n$ increases forever and diverges. Nothing about rising in an orderly way makes a sequence settle.
Bounded without monotone. $a_n = (-1)^n$ stays inside $[-1, 1]$ and diverges: the odd terms sit at $-1$ and the even at $+1$, and no single number is approached. A subtler version is $a_n = \sin n$, which is bounded, non-repeating, and wanders densely through $[-1,1]$ forever.
What the two together buy is a trap. Rising means the terms cannot come back; bounded means they cannot leave. The only thing left is to converge — and the number they converge to is the supremum, which completeness supplies.
That is the theorem's real value. Convergence is usually established by naming the limit and checking the definition, which requires knowing the answer. Here the answer is produced by the theorem. Every existence statement in the rest of this unit — that a bounded increasing sequence of partial sums has a sum, that a power series has a radius — ultimately rests on this one.
Treating a bounded sequence as convergent. $(-1)^n$ is the standing counterexample. Bounded rules out one kind of failure, not all of them.
Solving $L = f(L)$ before proving convergence exists. The equation has a solution for plenty of divergent recursions, and reporting it is asserting a limit that is not there.
Using L'Hopital on a sequence directly. The rule is about differentiable functions of a real variable. Pass to $f(x)$, apply it there, and inherit the answer.
Concluding divergence from a non-existent limit of the matching function. $\sin(\pi n)$ is identically $0$; $\sin(\pi x)$ has no limit. The implication runs one way only.
Confusing the sequence with the series. $a_n = \dfrac1n$ converges, to $0$. The series $\sum \dfrac1n$ diverges. These are different objects, and the next lesson depends on keeping them apart.
Worrying about the first few terms. Convergence is a property of the tail. A sequence that is increasing only from $n = 50$ onwards is covered by the theorem.
Every limit met so far was found by computing: divide by the dominant term, apply a rule, get a formula. So the monotone convergence theorem reads as a strange sort of statement — it concludes that a limit exists while offering no way to find it, which can feel like a theorem that has not finished.
It has. Existence is the hard part, and it is what everything later needs. The sum of an infinite series is defined as the limit of its partial sums, and for a series of positive terms those partial sums are increasing; so the entire question of whether such a series has a sum is the question of whether the partial sums are bounded. Every comparison test in the next four lessons is that observation applied.
The same pattern explains why the theorem is equivalent to completeness. Over the rationals, $1, 1.4, 1.41, 1.414, \ldots$ is increasing and bounded above by $2$, and it converges to nothing — the number it should converge to is missing from the system. The theorem is true of the reals precisely because the reals have no such gaps, and it is the form in which completeness is used for the rest of this course.
So when you meet a recursion, do it in the order the theorem imposes: prove it is monotone, prove it is bounded, conclude it converges, and only then find what to. The last step is algebra and the first three are the mathematics.
$a_1 = 1$, $a_{n+1} = \sqrt{2 + a_n}$. First: is it increasing? $a_2 = \sqrt3 > 1 = a_1$, and if $a_{n} > a_{n-1}$ then $\sqrt{2 + a_n} > \sqrt{2 + a_{n-1}}$, so yes by induction.
Monotone, by induction.
Bounded? If $a_n < 2$ then $a_{n+1} = \sqrt{2 + a_n} < \sqrt4 = 2$, and $a_1 < 2$. So every term is below $2$, again by induction.
Bounded, by induction.
Now the theorem applies, so a limit $L$ exists. Only now solve: $L = \sqrt{2 + L}$, so $L^2 - L - 2 = 0$, giving $L = 2$ or $L = -1$; the terms are positive, so $L = 2$.
The value comes last, and the existence first.
$a_n = \dfrac{\ln n}{n}$. L'Hopital does not apply to a sequence, so consider $f(x) = \dfrac{\ln x}{x}$ instead.
Pass to the real variable first.
$\dfrac{\ln x}{x}$ is $\dfrac{\infty}{\infty}$, and L'Hopital gives $\dfrac{1/x}{1} \to 0$.
The rule applies to $f$, which is differentiable.
Since $f(x) \to 0$ and $a_n = f(n)$, the sequence tends to $0$. The general fact behind it is that $\ln n$ loses to every positive power of $n$, so $\dfrac{\ln n}{n^{0.001}} \to 0$ as well.
Write it out: $\dfrac{n!}{n^n} = \dfrac{1}{n}\cdot\dfrac{2}{n}\cdots\dfrac{n}{n}$, a product of $n$ factors each at most $1$.
Look at the structure rather than reaching for a rule.
Every factor after the first is at most $1$, so $0 < a_n \le \dfrac1n$.
A bound that is easy to see and enough.
By the squeeze theorem, $a_n \to 0$. So $n!$ loses to $n^n$, which is the last comparison in the growth-rate chain.
Find $\displaystyle\lim_{n\to\infty}\frac{9n + 3}{4n + 9}$.
Answer:
Match each sequence to its limit.
| $0$ | $e$ | $4$ | |
|---|---|---|---|
| $a_n = \dfrac{\ln n}{n}$ | |||
| $a_n = \dfrac{n^{4}}{5^{\,n}}$ | |||
| $a_n = \left(1 + \dfrac{1}{n}\right)^{n}$ | |||
| $a_n = \dfrac{4n^{2} + 1}{n^{2} + 4}$ |
A sequence starts at $a_1 = 4$ and satisfies $a_{n+1} = \dfrac{a_n + 1}{2}$. Fill in the next three terms.
| Value | |
|---|---|
| $a_2$ | |
| $a_3$ | |
| $a_4$ |
Build the proof that an increasing sequence bounded above by $5$ converges.
This task has no paper form; do it on a device.
What does the sequence $a_n = (-1)^{n}\left(7 - \dfrac{1}{n}\right)$ do?
A sequence satisfies $a_{n+1} = \dfrac{a_n + 8}{6}$ and is known to converge. What is its limit?
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Find $\displaystyle\lim_{n\to\infty}\frac{6n + 3}{8n + 9}$.
Answer:
You can decide whether a sequence converges and find its limit, and say why bounded alone is not enough. Say in your own words what the monotone convergence theorem gives you that a computation cannot. Next: adding a sequence up.
10. Your turn: does $a_n = \dfrac{n!}{n^{n}}$ converge, and to what?, step 3