Chapter 1 — Sequences

By the end of this chapter you will:

1. Distinguish a sequence from a series
2. Decide whether a sequence converges and find its limit
3. Use the Monotonic Bounded Theorem (MBT) to prove convergence
4. Recognize the standard sequence patterns AP loves to test

1.0 Sequence vs. Series — One-Line Distinction

🔑 Key insight: A sequence converging does not imply the corresponding series converges. Example: $a_n = \frac{1}{n}$ converges to 0, but the harmonic series $\sum \frac{1}{n}$ diverges to $\infty$. This is one of BC's favorite traps.

🎵 Audio 1 — Sequence vs. Series intuition (Chinese narration) — audio/01_intro.mp3

1.1 Sequences — Intuitive Layer

Picture this: you are hopping along a number line. Each hop lands at a point $a_n$.

• If your landings cluster closer and closer to a fixed value $L$ → the sequence converges, $\lim_{n\to\infty} a_n = L$
• If your landings wander or run off to infinity → it diverges

Formal definition (good to recognize, intuition is enough day-to-day):
$$\lim_{n\to\infty} a_n = L \iff \forall \varepsilon > 0,\ \exists N,\ n>N \Rightarrow |a_n - L| < \varepsilon$$

In plain English: "no matter how small a tolerance $\varepsilon$ you pick, eventually every $a_n$ falls inside an $\varepsilon$-neighborhood of $L$."

1.2 Computing Limits — The Four Tools

Tool 1: Direct simplification

$$a_n = \frac{3n^2 + 2n}{n^2 + 5}$$

Divide top and bottom by the highest power $n^2$:

$$a_n = \frac{3 + 2/n}{1 + 5/n^2} \xrightarrow{n\to\infty} \frac{3+0}{1+0} = 3$$

Rule: "compare degrees."

• numerator degree > denominator → diverges to $\pm\infty$
• equal degrees → ratio of leading coefficients
• numerator degree < denominator → converges to 0

Tool 2: Replace $n$ by $x$, then L'Hôpital

$$a_n = \frac{\ln n}{n}$$

Treat $n$ as a continuous variable: $\lim_{x\to\infty}\frac{\ln x}{x} = \lim \frac{1/x}{1} = 0$.

⚠️ Trap: L'Hôpital only applies to $\frac{0}{0}$ or $\frac{\infty}{\infty}$. Verify the form first. Most AP problems are written in a form where this is legal.

Tool 3: Squeeze Theorem

When $a_n$ contains $\sin, \cos$, or $(-1)^n$ but is bounded between two known limits.

Example: $a_n = \frac{\sin n}{n}$.

Since $-1 \le \sin n \le 1$:
$$-\frac{1}{n} \le \frac{\sin n}{n} \le \frac{1}{n}$$

Both outer bounds approach 0, so $a_n \to 0$.

Tool 4: Memorized limits

🎵 Audio 4 — Four-tool overview — audio/04_four_tools.mp3

1.3 Monotonic Bounded Theorem (MBT)

Intuition: an elevator that only goes up (monotonically increasing) but is capped by a ceiling (bounded above) must stop at some floor.

Theorem:

• A sequence that is monotonically increasing AND bounded above converges.
• A sequence that is monotonically decreasing AND bounded below converges.

How to prove monotonicity — three strategies:

1. Sign of $a_{n+1} - a_n$
2. Compare $\frac{a_{n+1}}{a_n}$ with 1 (when all terms are positive)
3. Replace $n$ with $x$ and check derivative sign

How to find the bound — usually plug in a few terms, guess, then verify by induction.

🎵 Audio 2 — Elevator analogy + solution playbook — audio/02_mbt.mp3

1.4 Worked Examples (Standard → Twist → AP-Level)

Example 1 (standard)

Find $\displaystyle\lim_{n\to\infty} \frac{4n^3 - n}{2n^3 + n^2 + 1}$.

Solution: divide by $n^3$:
$$= \lim \frac{4 - 1/n^2}{2 + 1/n + 1/n^3} = \frac{4}{2} = 2$$

Converges to 2. ✅

Example 2 (twist)

Decide whether $a_n = \frac{n!}{2^n}$ converges.

Intuition: factorial grows faster than any exponential, so this should diverge.

Verify: $\frac{a_{n+1}}{a_n} = \frac{(n+1)!/2^{n+1}}{n!/2^n} = \frac{n+1}{2}$.

For $n \ge 2$, the ratio exceeds 1, so $a_n$ keeps growing — diverges to $\infty$.

🔑 Speed hierarchy (slowest to fastest):
$$\log n \prec n^p \prec a^n \ (a>1) \prec n! \prec n^n$$
Whichever is in the numerator and slower → limit is 0. In the denominator and slower → limit is $\infty$.

Example 3 (AP-level — recursive sequence)

Let $a_1 = 1$ and $a_{n+1} = \sqrt{2 + a_n}$. Show $\{a_n\}$ converges and find its limit.

Step 1 — Monotonicity? Compute a few terms: $a_1=1,\ a_2=\sqrt{3}\approx 1.73,\ a_3 \approx 1.93$. Looks increasing.

Induction: assume $a_n > a_{n-1}$. Then $2 + a_n > 2 + a_{n-1}$, so $\sqrt{2+a_n} > \sqrt{2+a_{n-1}}$, i.e. $a_{n+1} > a_n$. ✓

Step 2 — Bounded above? Guess upper bound $= 2$. Induction: $a_n < 2 \Rightarrow 2 + a_n < 4 \Rightarrow a_{n+1} = \sqrt{2+a_n} < 2$. ✓

By MBT, the sequence converges.

Step 3 — Find $L$. Let $L = \lim a_n$. Take limits on both sides of the recursion:
$$L = \sqrt{2 + L} \Rightarrow L^2 - L - 2 = 0 \Rightarrow L = 2 \text{ or } -1$$

Since $a_n > 0$, $L = 2$. ✅

🔑 AP playbook for recursive sequences: prove MBT → set $L$ → solve the equation.

🎵 Audio 5 — Recursive-sequence three-step walkthrough — audio/05_example3.mp3

1.5 Trap Alerts ⚠️

1. Sequence convergence ≠ series convergence. $\frac{1}{n} \to 0$ but $\sum \frac{1}{n}$ diverges.
2. Verify the L'Hôpital form first. Only $\frac{0}{0}$ or $\frac{\infty}{\infty}$ qualify.
3. $(-1)^n$ alone diverges — it bounces between $\pm 1$. But $\frac{(-1)^n}{n} \to 0$ via Squeeze.
4. A "candidate" limit is not the limit until existence is proved. In Example 3 you must do MBT first, then solve. Don't skip.

🎵 Audio 6 — Four trap alerts — audio/06_traps.mp3

1.6 Mnemonic

"Compare degrees, remember $e$ and $1$, squeeze for noise, MBT for recursion."

• Compare degrees: rational expressions reduce to leading-coefficient ratio
• $e$ and $1$: $(1+1/n)^n \to e$, $\sqrt[n]{\cdot} \to 1$
• Squeeze: handles $\sin, \cos, (-1)^n$
• MBT: handles recursive definitions

🎵 Audio 3 — Recap + mnemonic — audio/03_recap.mp3

Media Inventory

End of chapter. Next: series basics, geometric series, and p-series.