Sequences: When Lists Have Limits

20.1 A New Kind of Function, and Why Part IV Begins Here

For nineteen chapters you have studied functions of a continuous variable: $f(x)$ where $x$ can be any real number, sliding smoothly along the line. Calculus has been the study of what happens as $x$ flows toward a point or toward infinity. Now we change one thing — we let the input march in discrete steps $1, 2, 3, \ldots$ instead of flowing — and an entire new world of calculus opens up.

A sequence is a function whose domain is $\mathbb{N}$, the natural numbers. Instead of writing $f(n)$ we write $a_n$ for the value at index $n$, and we write $\{a_n\}_{n=1}^\infty$ — or just $\{a_n\}$ — for the whole infinite list. Nothing more exotic is going on: a sequence is a rule that hands you a real number for each counting number.

$$a_1, \ a_2, \ a_3, \ a_4, \ \ldots$$

Here are five sequences worth meeting at once, because each foreshadows a behavior we will spend the chapter pinning down.

The Key Insight. A sequence is just a function sampled at the integers, and the limit of a sequence is the limit-at-infinity of Chapter 3 with the continuous neighborhood replaced by a cutoff index $N$. Everything you learned about limits transfers — you only have to translate "$x$ near $a$" into "$n$ at least $N$."

Why this is the opener of Part IV. Part IV is about infinite processes done correctly: adding infinitely many numbers (series, Chapter 21), testing whether such sums converge (Chapter 22), and building functions out of infinite polynomials (power and Taylor series, Chapter 23), culminating in applications like Euler's identity (Chapter 24). Every one of those is secretly a statement about a sequence. A series is nothing but the sequence of its partial sums; a Taylor series converges exactly when a sequence of approximations converges. Master sequences here and the rest of Part IV becomes the study of which sequences converge and how fast. This is the shortest topic in Part IV, but it is the floor everything else stands on.

Geometric Intuition. Picture the graph of a sequence as a scatter of dots above the integers $n = 1, 2, 3, \ldots$ on a horizontal axis. Convergence means the dots eventually crowd into a horizontal band of any width you like, hugging a line $y = L$. Divergence means they refuse to settle — they march off to infinity, or they hop back and forth forever, or they wander. The whole chapter is about reading these dot-pictures precisely.

20.2 Three Ways to Define a Sequence

A sequence can be handed to you in several disguises, and recognizing them is half the battle.

Explicit formula. A rule for $a_n$ directly in terms of $n$. Example: $a_n = \dfrac{2n+1}{n+3}$. You can jump straight to the 1000th term without computing the first 999.

Recursive formula. Each term is built from earlier ones, plus a starting value (or values). Example: $a_1 = 1$, $a_{n+1} = \tfrac12\!\left(a_n + \dfrac{2}{a_n}\right)$ — the Babylonian iteration for $\sqrt{2}$ we analyze in §20.6. Recursive sequences are how algorithms generate numbers, and they will be the deepest part of this chapter.

Implicit enumeration. $a_n$ is "the $n$-th object" in some natural list — the $n$-th prime, say: $2, 3, 5, 7, 11, \ldots$. There is no formula, only a procedure. These are common in number theory but rare in calculus; we mention them only so you recognize a sequence when you see one.

The explicit and recursive forms dominate calculus, and they pull in opposite computational directions: explicit formulas are easy to evaluate but can be hard to discover, while recursive formulas are easy to discover but can be hard to evaluate in closed form. The Fibonacci numbers (recursive, simple) and Binet's formula (explicit, surprising) for the same sequence dramatize the trade-off — we return to it in §20.7.

Check Your Understanding. Write the first four terms of the sequence $a_n = \dfrac{(-1)^n}{2^n}$, and the first four terms of the recursive sequence $b_1 = 3$, $b_{n+1} = \tfrac12 b_n + 1$.

Answer

For $a_n$: $a_1 = -\tfrac12$, $a_2 = \tfrac14$, $a_3 = -\tfrac18$, $a_4 = \tfrac{1}{16}$ — alternating in sign, shrinking toward $0$. For $b_n$: $b_1 = 3$, $b_2 = \tfrac12(3)+1 = 2.5$, $b_3 = \tfrac12(2.5)+1 = 2.25$, $b_4 = 2.125$ — decreasing toward $2$ (the fixed point of $f(x) = \tfrac12 x + 1$, since $x = \tfrac12 x + 1 \Rightarrow x = 2$).

20.3 The Limit of a Sequence

We now make "settles down to $L$" precise. This is the central definition of the chapter, and it is the ε-δ definition of Chapter 3 wearing new clothes.

A sequence $\{a_n\}$ converges to the number $L$, written

$$\lim_{n \to \infty} a_n = L \qquad\text{or}\qquad a_n \to L,$$

if for every $\varepsilon > 0$ there exists an integer $N$ such that

$$n \geq N \implies |a_n - L| < \varepsilon.$$

Read it as a challenge–response game. An adversary names a tolerance $\varepsilon$ — a band of half-width $\varepsilon$ around $L$. You must produce a cutoff $N$ such that every term from the $N$-th onward lands inside the band. If you can always answer, no matter how small $\varepsilon$ gets, the sequence converges to $L$.

A sequence that converges to some finite $L$ is convergent. A sequence that does not is divergent — it may diverge to $\pm\infty$ (like $n!$), oscillate forever (like $(-1)^n$), or wander erratically.

Worked Example 20.3.1 — proving $1/n \to 0$ from the definition. Let $a_n = 1/n$ and $L = 0$. Given $\varepsilon > 0$, we need $|1/n - 0| = 1/n < \varepsilon$, i.e. $n > 1/\varepsilon$. So choose $N$ to be any integer larger than $1/\varepsilon$. Then for all $n \geq N$ we have $n > 1/\varepsilon$, hence $1/n < \varepsilon$. The challenge is met for every $\varepsilon$, so $1/n \to 0$. For instance, if $\varepsilon = 0.001$, take $N = 1001$; every term from $a_{1001}$ on lies within $0.001$ of $0$.

Geometric Intuition. The $\varepsilon$-band is a horizontal strip of half-width $\varepsilon$ centered on $y = L$. Convergence says: no matter how thin you make the strip, only finitely many dots stick out of it; all the rest — an infinite tail — lie inside. The index $N$ is just "where the dots have finished entering the strip for good."

Math Major Sidebar — How ε-N is the same animal as ε-δ. In Chapter 3, $\lim_{x\to\infty} f(x) = L$ meant: for every $\varepsilon > 0$ there is an $M$ with $x > M \implies |f(x) - L| < \varepsilon$. The sequence definition is literally this statement with the continuous threshold $M$ replaced by an integer threshold $N$ and $x$ restricted to integers. Even the finite-point limit $\lim_{x\to a} f(x) = L$ shares the skeleton: there, $\delta$ controls how close the input must be; here, $N$ controls how far out the index must be. In both cases $\varepsilon$ controls how close the output must be. The deep payoff is the sequential characterization of limits: $f$ is continuous at $a$ if and only if $f(a_n) \to f(a)$ for every sequence $a_n \to a$. Continuity, the cornerstone of Chapter 4, can be defined entirely in the language of sequences — which is why analysts often build calculus on sequences from the ground up.

Sequences versus their continuous parents. If a function $f$ satisfies $\lim_{x\to\infty} f(x) = L$, then the sampled sequence $a_n = f(n)$ also converges to $L$ — you are just reading off the function at integer inputs. This is enormously useful: it lets you apply L'Hôpital's rule (Chapter 9) to a continuous version of a sequence. But the converse fails. The sequence $a_n = \sin(\pi n) = 0$ for all integers $n$ converges to $0$, while the function $\sin(\pi x)$ has no limit at infinity — the sequence simply never samples the oscillation. Sequences can be better-behaved than the functions they come from, because they only look at integer points.

Common Pitfall. Many students write "$(-1)^n \to$ something" because the terms stay bounded between $-1$ and $1$. But bounded is not convergent. The terms must approach a single value $L$. Since $(-1)^n$ revisits $-1$ and $+1$ forever, no single $L$ can trap an infinite tail inside a band of half-width, say, $\varepsilon = 0.5$. The sequence diverges by oscillation. Settling near one number is the whole point.

20.4 Limit Laws for Sequences

Because sequence limits are just limits, they inherit the algebra of limits from Chapter 3. Suppose $\lim a_n = A$ and $\lim b_n = B$, with $A$ and $B$ finite. Then:

$$\lim (a_n + b_n) = A + B, \qquad \lim (a_n - b_n) = A - B, \qquad \lim (c\,a_n) = cA,$$ $$\lim (a_n b_n) = AB, \qquad \lim \frac{a_n}{b_n} = \frac{A}{B}\ \ (B \neq 0), \qquad \lim |a_n| = |A|.$$

To these add the Continuous-Function Rule: if $g$ is continuous at $L$ and $a_n \to L$, then $g(a_n) \to g(L)$. This is the sequential characterization of continuity (§20.3 sidebar) put to work — it lets you pass a limit inside a continuous function. For example, since $a_n = 1 + 1/n \to 1$ and $g(x) = \sqrt{x}$ is continuous at $1$, we get $\sqrt{1 + 1/n} \to \sqrt{1} = 1$.

Worked Example 20.4.1 — a rational sequence. Find $\displaystyle\lim_{n\to\infty} \frac{n^2 + 3n}{2n^2 - 5}$. The leading powers tie at $n^2$, so divide numerator and denominator by $n^2$:

$$\frac{n^2 + 3n}{2n^2 - 5} = \frac{1 + 3/n}{2 - 5/n^2} \longrightarrow \frac{1 + 0}{2 - 0} = \frac{1}{2}.$$

Every "$1/n$" and "$1/n^2$" goes to $0$ by §20.3, and the limit laws assemble the pieces. The limit is $\tfrac12$ — the ratio of leading coefficients, exactly as your intuition about "highest-degree wins" predicts.

The Squeeze Theorem for Sequences

When a sequence is tangled up with an oscillation, the limit laws alone may not reach it. The Squeeze Theorem (also called the Sandwich Theorem) does.

Squeeze Theorem for Sequences. If $b_n \leq a_n \leq c_n$ for all $n$ beyond some index, and $b_n \to L$ and $c_n \to L$, then $a_n \to L$.

The middle sequence has no room to do anything but converge to $L$: it is pinned between two sequences that both head there.

Worked Example 20.4.2 — taming an oscillation. Find $\displaystyle\lim_{n\to\infty} \frac{\sin n}{n}$. The numerator oscillates unpredictably, so no formula-manipulation reaches the limit. But $-1 \leq \sin n \leq 1$, so dividing by the positive quantity $n$,

$$-\frac{1}{n} \leq \frac{\sin n}{n} \leq \frac{1}{n}.$$

Both outer sequences converge to $0$, so the squeeze forces $\dfrac{\sin n}{n} \to 0$. The wild oscillation of $\sin n$ is crushed by the shrinking envelope $\pm 1/n$.

Real-World Application — Signal averaging in engineering. A noisy sensor reading at sample $n$ can be modeled as $a_n = L + \frac{\text{noise}_n}{n}$, where the noise term is bounded (say $|\text{noise}_n| \le M$) but erratic. The Squeeze Theorem guarantees $a_n \to L$: bounded noise divided by a growing number of accumulated samples is annihilated. This is the mathematical heart of why averaging more measurements sharpens an estimate — and a baby version of the Law of Large Numbers, where sample means of independent measurements converge to the true value.

20.5 A Catalog of Standard Limits

A handful of limits recur so often — especially in the convergence tests of Chapter 22 — that they are worth memorizing. Each can be proved from the definition or from a continuous analog via L'Hôpital, but in practice you cite them.

Behind the table is a single picture — a hierarchy of growth rates. As $n \to \infty$, each of the following overwhelms the one before it:

$$\ln n \ \prec\ n^p \ \prec\ r^n \ \prec\ n! \ \prec\ n^n \qquad (p > 0,\ r > 1).$$

Logarithms crawl, powers walk, exponentials run, factorials sprint, and $n^n$ outruns even factorials. Whenever you see a ratio of two of these, the faster-growing one wins: it dominates the limit. This single ordering decides most of the convergence questions in Chapters 21 and 22.

Worked Example 20.5.1 — factorial beats power. Find $\displaystyle\lim_{n\to\infty} \frac{n!}{n^n}$. Write the ratio as a product of $n$ fractions:

$$\frac{n!}{n^n} = \frac{n}{n}\cdot\frac{n-1}{n}\cdot\frac{n-2}{n}\cdots\frac{1}{n}.$$

Every factor is $\leq 1$, and the last factor is exactly $1/n$. So the whole product is at most $1/n$:

$$0 < \frac{n!}{n^n} \leq \frac{1}{n} \longrightarrow 0.$$

By the Squeeze Theorem (lower bound $0$, upper bound $1/n$), the limit is $0$. Even though both $n!$ and $n^n$ explode, $n^n$ explodes faster, and the ratio collapses to $0$ — exactly what the hierarchy predicts.

Worked Example 20.5.2 — the compound-interest limit. Find $\displaystyle\lim_{n\to\infty}\left(1 + \frac{2}{n}\right)^{\!n}$. This is the standard limit $\left(1 + x/n\right)^n \to e^x$ with $x = 2$, so the answer is $e^{2}$. We will derive this limit from scratch and connect it to money in §20.8 — it is one of the most important sequence limits in all of applied mathematics.

Check Your Understanding. Evaluate $\displaystyle\lim_{n\to\infty} \frac{3^n}{n!}$ using the hierarchy, and $\displaystyle\lim_{n\to\infty} n^{1/n}$.

Answer

For $3^n/n!$: in the hierarchy $r^n \prec n!$, so the factorial dominates and the ratio $\to 0$. (Intuitively, once $n > 3$ every new factor in $n!$ exceeds the new factor $3$ in $3^n$, so the denominator pulls ahead permanently.) For $n^{1/n}$: this is the tabled standard limit, equal to $1$. Quick check via logs: $\ln(n^{1/n}) = (\ln n)/n \to 0$ since $\ln n \prec n$, and $e^0 = 1$.

20.6 The Monotone Convergence Theorem

So far, proving convergence has meant finding the limit. But the most powerful convergence tool requires you to know almost nothing about the limit — only that the sequence climbs (or descends) and cannot escape to infinity.

Call a sequence monotone if it is either non-decreasing ($a_{n+1} \geq a_n$ for all $n$) or non-increasing ($a_{n+1} \leq a_n$ for all $n$). Call it bounded above if some $M$ satisfies $a_n \leq M$ for all $n$, and bounded below if some $m$ satisfies $a_n \geq m$.

Monotone Convergence Theorem (MCT). A bounded monotone sequence converges. More precisely: - non-decreasing and bounded above $\Rightarrow$ converges to its least upper bound (supremum); - non-increasing and bounded below $\Rightarrow$ converges to its greatest lower bound (infimum).

Geometric Intuition. Imagine the dots climbing steadily to the right but never crossing a ceiling line $y = M$. They cannot rise forever (the ceiling stops them) and they cannot fall back (monotone forbids it). The only option left is to pile up against some height $L \le M$ — and that height is the supremum. The theorem is the mathematical statement of "what goes up and is capped must level off." Its truth depends on the completeness of the real numbers: $\mathbb{R}$ has no gaps for the limit to fall into.

The MCT's superpower is that you don't need to know $L$ in advance. You prove monotonicity and boundedness — two often-checkable facts — and convergence comes for free. This is exactly the situation for recursive sequences, where a closed form for $L$ may be hard to find but monotonicity and bounds are within reach.

Worked Example 20.6.1 — The Babylonian Method for $\sqrt{2}$

Define the recursive sequence

$$a_1 = 1, \qquad a_{n+1} = \frac{1}{2}\!\left(a_n + \frac{2}{a_n}\right).$$

Compute a few terms:

$$a_1 = 1,\quad a_2 = \tfrac12(1 + 2) = 1.5,\quad a_3 = \tfrac12(1.5 + \tfrac{2}{1.5}) \approx 1.41667,\quad a_4 \approx 1.414216,\ \ldots$$

The terms appear to race toward $\sqrt{2} = 1.41421356\ldots$. Let us prove convergence with the MCT, in three moves.

Step 1 — bounded below by $\sqrt 2$ (for $n \ge 2$). For any $x > 0$, the AM–GM inequality gives $\tfrac12(x + \tfrac{2}{x}) \geq \sqrt{x \cdot \tfrac{2}{x}} = \sqrt{2}$. Applying this with $x = a_n$ shows $a_{n+1} \geq \sqrt{2}$. So from the second term onward, every term is at least $\sqrt 2$; the sequence is bounded below.

Step 2 — non-increasing (for $n \ge 2$). Once $a_n \geq \sqrt 2$, we have $a_n^2 \geq 2$, hence $\dfrac{2}{a_n} \leq a_n$. Averaging $a_n$ with the smaller number $2/a_n$ cannot increase it:

$$a_{n+1} = \frac{1}{2}\!\left(a_n + \frac{2}{a_n}\right) \leq \frac{1}{2}(a_n + a_n) = a_n.$$

So for $n \geq 2$ the sequence is non-increasing.

Step 3 — apply the MCT and find the limit. The tail $a_2, a_3, a_4, \ldots$ is non-increasing and bounded below by $\sqrt 2$, so by the MCT it converges to some limit $L \geq \sqrt 2$. Because the recursion $a_{n+1} = \tfrac12(a_n + 2/a_n)$ is continuous in $a_n$, taking $n \to \infty$ on both sides (using the Continuous-Function Rule of §20.4) gives the fixed-point equation

$$L = \frac{1}{2}\!\left(L + \frac{2}{L}\right) \;\Longrightarrow\; 2L = L + \frac{2}{L} \;\Longrightarrow\; L = \frac{2}{L} \;\Longrightarrow\; L^2 = 2.$$

Since $L \geq \sqrt 2 > 0$, we conclude $L = \sqrt 2$. The Babylonian iteration converges to $\sqrt 2$ — a 3,500-year-old algorithm validated by a 19th-century theorem.

Historical Note. This iteration appears on Babylonian clay tablets (notably YBC 7289, c. 1800–1600 BCE), where $\sqrt 2$ is recorded to the equivalent of six decimal places. The method was rediscovered by Heron of Alexandria (c. 60 CE) and is now recognized as Newton's method applied to $f(x) = x^2 - 2$ (Chapter 11). Three civilizations found the same recursion because it is, in a precise sense, the natural one — a theme we make explicit in §20.7.

20.7 Recursive Sequences and Fixed Points

The Babylonian example is a special case of a pattern that runs through all of computational mathematics: a recursion of the form

$$a_{n+1} = f(a_n).$$

If such a sequence converges to a limit $L$, and $f$ is continuous, then taking the limit of both sides forces

$$L = f(L).$$

A value $a^{*}$ satisfying $f(a^{*}) = a^{*}$ is called a fixed point of $f$ — a point that $f$ leaves unchanged. So the only possible limits of $a_{n+1} = f(a_n)$ are the fixed points of $f$. Finding limits of recursive sequences reduces to solving the equation $f(x) = x$.

But fixed points come in two flavors, and only one attracts nearby sequences.

Stability Criterion. Let $a^{*}$ be a fixed point of a differentiable $f$. Then $a^{*}$ is attracting (locally stable) if $|f'(a^{*})| < 1$: iteration starting near $a^*$ converges to $a^*$. It is repelling (unstable) if $|f'(a^{*})| > 1$: iteration is pushed away. When $|f'(a^{*})| = 1$, the test is inconclusive and behavior is delicate.

Why the criterion holds. Near the fixed point, linearize $f$ using the tangent-line approximation of Chapter 11: $f(x) \approx f(a^*) + f'(a^*)(x - a^*) = a^* + f'(a^*)(x - a^*)$. Subtracting $a^*$, the error $e_n = a_n - a^*$ evolves as

$$e_{n+1} = a_{n+1} - a^{*} \approx f'(a^{*})\, e_n.$$

So each step multiplies the error by roughly $f'(a^*)$. If $|f'(a^*)| < 1$ the error shrinks geometrically to $0$ (convergence); if $|f'(a^*)| > 1$ it amplifies (divergence). The stability of a numerical algorithm is, quite literally, a derivative.

Worked Example 20.7.1 — Iterating the Cosine

Press the cosine button on a calculator (in radians) over and over, starting from $x_0 = 1$:

$$1,\ 0.5403,\ 0.8576,\ 0.6543,\ 0.7935,\ 0.7014,\ \ldots$$

The values oscillate but visibly close in on a single number. They are the recursive sequence $x_{n+1} = \cos(x_n)$, so the limit must be a fixed point of $\cos$: a solution of $x = \cos x$. That solution is the Dottie number, $x^{*} \approx 0.7390851$. Checking stability with $f(x) = \cos x$, $f'(x) = -\sin x$:

$$|f'(x^{*})| = |\sin(0.7391)| \approx 0.6736 < 1,$$

so the fixed point is attracting — the iteration must converge, exactly as observed. The sign of $f'$ is negative, which is why the approach oscillates (each step overshoots and the error flips sign) rather than creeping in from one side.

import numpy as np

# Fixed-point iteration x_{n+1} = cos(x_n), starting from x_0 = 1
x = 1.0
for i in range(20):
    x = np.cos(x)
    if i < 4 or i > 17:
        print(f"x_{i+1} = {x:.10f}")

# x_1 = 0.5403023059
# x_2 = 0.8575532158
# x_3 = 0.6542897905
# x_4 = 0.7934803587
# x_19 = 0.7390546704
# x_20 = 0.7390958919
# Settling toward the Dottie number x* = 0.7390851332...

Newton's Method Is a Recursive Sequence

The most important recursive sequence in applied mathematics is Newton's method from Chapter 11. To solve $g(x) = 0$, iterate

$$x_{n+1} = x_n - \frac{g(x_n)}{g'(x_n)} = f(x_n), \qquad f(x) = x - \frac{g(x)}{g'(x)}.$$

The fixed points of $f$ are exactly the roots of $g$: $f(x) = x \iff g(x)/g'(x) = 0 \iff g(x) = 0$. So Newton's iteration is a fixed-point sequence whose target is a root. Its stability is spectacular. Differentiating $f$ and simplifying gives $f'(x) = \dfrac{g(x)\,g''(x)}{g'(x)^2}$, which vanishes at a simple root (where $g = 0$, $g' \neq 0$). With $f'(a^*) = 0$, the error recursion $e_{n+1} \approx f'(a^*)e_n$ loses its linear term entirely, and a second-order expansion shows

$$|e_{n+1}| \approx C\, |e_n|^2 \qquad (\text{quadratic convergence}).$$

The number of correct digits roughly doubles each step — which is why $a_4$ already matched $\sqrt 2$ to five decimals in §20.6. The Babylonian method is Newton's method on $g(x) = x^2 - 2$: $$x_{n+1} = x_n - \frac{x_n^2 - 2}{2x_n} = \frac{1}{2}\!\left(x_n + \frac{2}{x_n}\right).$$ The 1800 BCE clay tablet and the 1669 De analysi manuscript hold the same recursion.

Common Pitfall. Solving the fixed-point equation $L = f(L)$ tells you the candidate limit, but it does not prove the sequence converges. A sequence sitting on a repelling fixed point flies away from it; a sequence may also diverge to infinity and have no finite limit at all. You must separately establish convergence — via the MCT, the stability criterion $|f'(a^*)| < 1$, or direct estimation — before you are entitled to substitute $L = f(L)$. Writing "the limit satisfies $L = f(L)$, therefore the sequence converges" reverses the logic.

Real-World Application — Root-finding everywhere in computation. Newton's method (and its cousins) is the engine inside almost every numerical solver: a calculator computing $\sqrt{2}$ or $\ln 7$, a financial model solving for a bond's yield-to-maturity, a graphics engine intersecting a ray with a surface, and the optimizers that train machine-learning models all run recursive sequences to convergence. The questions of this chapter — does the sequence converge, to what, and how fast — are the questions every numerical analyst asks before trusting an answer.

20.8 Application: Compound Interest and the Birth of $e$

Few sequences pay off as literally as the compound-interest sequence. Invest one dollar at annual rate $r$, compounded $n$ times per year. After one year the balance is

$$a_n = \left(1 + \frac{r}{n}\right)^{\!n}.$$

Compound annually ($n=1$), and you get $1 + r$. Compound monthly ($n = 12$), quarterly, daily — the balance grows as $n$ rises, because interest starts earning interest sooner. The natural question: is there a ceiling? What happens under continuous compounding, $n \to \infty$?

Take $r = 1$ (a generous 100% rate) and watch the sequence climb:

# Compound interest with r = 1 as compounding frequency n increases
for n in [1, 2, 12, 365, 8760, 525600]:
    print(f"n = {n:7d}:  (1 + 1/n)^n = {(1 + 1/n)**n:.10f}")

# n =       1:  (1 + 1/n)^n = 2.0000000000   (annual)
# n =       2:  (1 + 1/n)^n = 2.2500000000   (semiannual)
# n =      12:  (1 + 1/n)^n = 2.6130352902   (monthly)
# n =     365:  (1 + 1/n)^n = 2.7145674820   (daily)
# n =    8760:  (1 + 1/n)^n = 2.7181266916   (hourly)
# n =  525600:  (1 + 1/n)^n = 2.7182792154   (every minute)

The balances rise but are visibly converging — not exploding — toward a number near $2.71828$. That number is Euler's number $e$:

$$\lim_{n\to\infty}\left(1 + \frac{1}{n}\right)^{\!n} = e \approx 2.718281828\ldots$$

This is the standard limit from §20.5 with $x = 1$, and we can see why it converges with the MCT. One can show the sequence $a_n = (1 + 1/n)^n$ is increasing and bounded above by $3$ (the binomial expansion of each term is dominated term-by-term by the convergent geometric-type series for $e$). Increasing + bounded above $\Rightarrow$ convergent, by the MCT — and the limit is defined to be $e$. So $e$ is not an arbitrary constant; it is the answer to "what does compound interest approach when you compound infinitely often?" For general rate $r$, the same argument gives $(1 + r/n)^n \to e^r$, the continuous-compounding formula behind everything from savings accounts to radioactive decay to the exponential growth models of Chapter 19.

Geometric Intuition. Each increase in $n$ adds more "interest on interest," nudging the one-year balance up — but with diminishing returns, because the extra compounding events each contribute less. The increments shrink fast enough to leave a finite gap below the ceiling $e$. Going from annual to monthly buys you a lot; going from "every minute" to "every second" buys you almost nothing. That diminishing-returns curve flattening to $e$ is the MCT made visible in dollars.

20.9 Sequences Are the Foundation of Part IV

Step back and notice how much of what follows is secretly about sequences.

• A series $\sum_{k=1}^\infty a_k$ (Chapter 21) is defined as the limit of its sequence of partial sums $s_n = a_1 + a_2 + \cdots + a_n$. "The series converges" means "the sequence $\{s_n\}$ converges." Everything in this chapter applies directly.
• The convergence tests of Chapter 22 are, at bottom, theorems about sequences — the ratio test uses the standard limit of $|a_{n+1}/a_n|$; the integral test compares a sequence to an integral.
• Taylor series (Chapter 23) converge when a sequence of polynomial approximations converges to the target function, term by term.
• The completeness of $\mathbb{R}$ — that bounded monotone sequences have real limits (the MCT) — is what guarantees these infinite processes land on actual numbers rather than falling through gaps.

The Key Insight. Every infinite process in calculus is a sequence in disguise. A series is a sequence of partial sums; a Taylor approximation is a sequence of polynomials; a numerical method is a sequence of estimates. Convergence of the process is convergence of the underlying sequence. Once you can decide whether a sequence has a limit, you hold the master key to all of Part IV.

Warning — sequences are not series, and the distinction matters. A sequence is a list $a_1, a_2, a_3, \ldots$; a series is a sum $a_1 + a_2 + a_3 + \cdots$. The sequence $a_n = 1/n$ converges (to $0$), but the series $\sum 1/n$ — the harmonic series — diverges to infinity (Chapter 21). The terms of a series can shrink to zero while the running total grows without bound. Confusing "the terms go to zero" with "the sum is finite" is the single most common error in all of Part IV. This chapter studies the terms; Chapter 21 studies the sums.

20.10 Computation: Verifying Convergence Numerically

The three-tier pattern — analytic statement, hand computation, machine confirmation — applies to sequences cleanly. Here we verify the Babylonian limit of §20.6 and watch the quadratic convergence that §20.7 predicted.

import numpy as np

# Babylonian iteration for sqrt(2): a_{n+1} = (a_n + 2/a_n)/2
a = 1.0
target = np.sqrt(2)
for n in range(1, 7):
    a = 0.5 * (a + 2.0 / a)
    err = abs(a - target)
    print(f"a_{n+1} = {a:.15f}   |error| = {err:.2e}")

# a_2 = 1.500000000000000   |error| = 8.58e-02
# a_3 = 1.416666666666667   |error| = 2.45e-03
# a_4 = 1.414215686274510   |error| = 2.12e-06
# a_5 = 1.414213562374690   |error| = 1.59e-12
# a_6 = 1.414213562373095   |error| = 2.22e-16  (machine precision)
# a_7 = 1.414213562373095   |error| = 2.22e-16

Computational Note. Read the error column: $8.6\times 10^{-2} \to 2.4\times 10^{-3} \to 2.1\times 10^{-6} \to 1.6\times 10^{-12}$. Each error is roughly the square of the previous one ($(10^{-3})^2 \approx 10^{-6}$, $(10^{-6})^2 \approx 10^{-12}$) — the signature of quadratic convergence that §20.7 derived from $f'(a^*) = 0$. The correct-digit count doubles every step, and the sequence hits machine precision ($\approx 2\times 10^{-16}$) by the fifth term. This is why Newton-type iterations dominate numerical computing: a handful of steps reach the limit of what floating-point arithmetic can even represent.

You can equally confirm the symbolic limits with sympy, which evaluates sequence limits directly:

import sympy as sp

n = sp.symbols('n', positive=True)
print(sp.limit((1 + 1/n)**n, n, sp.oo))        # E   (Euler's number e)
print(sp.limit((n**2 + 3*n)/(2*n**2 - 5), n, sp.oo))  # 1/2  (Worked Example 20.4.1)
print(sp.limit(sp.factorial(n)/n**n, n, sp.oo))       # 0    (Worked Example 20.5.1)

The machine confirms by hand result and symbolic engine agree — hand computation built the understanding, and the code certifies it.

20.11 Summary and the Road Ahead

You have met the discrete side of calculus and the toolkit for deciding when an infinite list settles down.

• A sequence is a function $\mathbb{N} \to \mathbb{R}$, given explicitly, recursively, or by enumeration.
• It converges to $L$ ($a_n \to L$) if for every $\varepsilon > 0$ there is an $N$ with $n \ge N \Rightarrow |a_n - L| < \varepsilon$ — the ε-N twin of Chapter 3's ε-δ limit.
• Limit laws (sums, products, quotients, continuous functions) and the Squeeze Theorem compute most limits; a small catalog of standard limits plus the growth hierarchy $\ln n \prec n^p \prec r^n \prec n! \prec n^n$ handles the rest.
• The Monotone Convergence Theorem proves convergence from monotonicity + boundedness without knowing the limit — the key to recursive sequences.
• A recursive sequence $a_{n+1} = f(a_n)$ can only converge to a fixed point $L = f(L)$, and does so when $|f'(a^*)| < 1$. Newton's method is such a sequence, converging quadratically to a root.
• Sequences underlie series (Chapter 21), convergence tests (Chapter 22), and Taylor series (Chapter 23). A sequence is a list; a series is a sum — never confuse them.

Add to Your Modeling Portfolio. Add a convergent iteration to your model, and report its limit and how fast it gets there. Biology: model a population under a discrete logistic update $p_{n+1} = r\,p_n(1 - p_n)$; find the stable equilibrium $p^* = 1 - 1/r$ for $1 < r < 3$ and check $|f'(p^*)| < 1$. Economics: model a market converging to equilibrium price via a recursive adjustment $P_{n+1} = P_n - k\,(\text{excess demand})$; identify the fixed point and its stability. Physics: model a damped oscillator's successive amplitudes as a geometric sequence $A_{n+1} = \rho\,A_n$ with $|\rho| < 1$, and find the total settling behavior. Data Science: implement gradient descent $\theta_{n+1} = \theta_n - \eta\,g'(\theta_n)$ as a recursive sequence (the anchor example from Chapter 6), and use the stability criterion to explain why too large a learning rate $\eta$ makes it diverge — convergence here is exactly the fixed-point analysis of §20.7.

Looking ahead. Chapter 21 takes the decisive step from lists to sums: it defines a series as the limit of its partial-sum sequence and meets the geometric and harmonic series — the first showing a convergent sum, the second showing that shrinking terms need not sum to something finite. Chapter 22 builds the full battery of convergence tests for the hard cases. Chapter 23 reaches the summit — power and Taylor series — where infinite polynomials reconstruct $e^x$, $\sin x$, and the normal-curve integral we have been chasing since Chapter 13. Every one of those triumphs rests on the single question you can now answer: does this sequence converge?