# Chapter 22: Sequences and Series: Basic Concepts

> From convergent sequences to geometric growth: what convergence means, the tests series obey, and why the improper-integral playbook transfers wholesale.

**Canonical:** https://calculus.kitzo.work/chapter/sequences-series  
**Level:** Advanced  
**Estimated time:** 55 minutes  
**Exercises:** 8 (auto-marked; answers stay server-side)

## Topics
- Sequences: convergence
- Two famous sequences
- Geometric series
- nth-term test
- Comparison & p-tests
- Ratio & root tests
- Integral test
- Alternating series

**Why this matters**

Sequences and series are the discrete twin of improper integrals: instead of asking whether an endless area is finite, you ask whether an endless sum is. The same fingerprints — blow-ups, comparison partners, cancellation — show up again, which makes this chapter both new and familiar.

## 1 · Convergence of sequences — does it settle?

A **sequence** converges to L if its terms eventually stay within any ε > 0 of L, however tiny ε is; otherwise it **diverges**. The classic family: `aₙ = a/(a + n)` → 1, because n swamps the constant a. Same for n/(2n + 7) → ½: divide top and bottom by the highest power of n.

A sequence can wobble and still converge, and convergence concerns the *far* tail — the first million terms are irrelevant.

## 2 · The bridge to functions

Every sequence is f at integer points: aₙ = f(n). The connection runs both ways. If `limx→∞ f(x) = L` for continuous f, then `limn→∞ f(n) = L` too — squeeze the continuous limit and the discrete one follows. (The converse can fail: sin(πn) is 0 for every integer n, while sin(πx) oscillates forever along the reals.) So you may borrow l'Hôpital and limits at infinity from calculus.

## 3 · Two sequences worth memorizing

- **(1 + 1/n)ⁿ → e ≈ 2.71828**. The terms creep upward: n = 10 gives ≈ 2.594, n = 100 ≈ 2.705. This is the compound-interest limit and the launchpad for eˣ.

- **n1/n → 1**. n1/n = eln n / n and ln n / n → 0, so 10001/1000 ≈ 1.007.

## 4 · From sequences to series; the geometric engine

A **series** Σ aₙ asks whether the running totals SN = a₁ + … + aₙ settle. The model citizen is the **geometric series**: `Σn=0∞ arn = a/(1 − r)`, and this holds **if and only if |r| < 1**./(1 − r); as N → ∞, rN+1 dies exactly when |r| < 1.) If |r| ≥ 1 the terms do not even shrink to zero, so the series diverges. Sums from a later starting index k work the same way: Σn=k∞ arn = ark/(1 − r).

## 5 · The nth-term test — the cheapest kill in the toolkit

**If aₙ ↛ 0, then Σ aₙ diverges.** Nothing else needs checking. A series of terms that refuse to die can never add to a finite number. The converse is false: aₙ → 0 does *not* rescue you — the harmonic series Σ 1/n diverges even though 1/n → 0. The nth-term test only ever proves divergence, never convergence.

## 6 · The improper-integral parallel games

Positive-term series obey the same tests as improper integrals — the analogy is deliberate:

- **Comparison**: 0 ≤ aₙ ≤ bₙ and Σbₙ converges ⇒ Σaₙ converges.

- **Limit comparison**: lim aₙ/bₙ = L ∈ (0, ∞) ⇒ same fate for both series.

- **p-test**: Σ 1/np converges iff p > 1. (The integral test proves it: ∫1∞ dx/xp and Σ 1/np share their fate exactly when f(x) = 1/xp is positive and shrinking.)

- **Absolute convergence**: Σ|aₙ| converges ⇒ Σaₙ converges.

**Ratio test**: let L = lim |aₙ₊₁/aₙ|. Then L < 1 ⇒ converges absolutely; L > 1 (or = ∞) ⇒ diverges; L = 1 ⇒ silent — no verdict. Shines on factorials and exponentials.

**Root test**: let L = lim |aₙ|1/n, same trichotomy (L < 1 converge, L > 1 diverge, L = 1 inconclusive). It shines when the nth power is baked into the term.

**Integral test**: for positive, decreasing, continuous f, Σ f(n) and ∫1∞ f(x) dx share fate — and Σn>N f(n) ≤ ∫N∞ f(x) dx bounds the tail.

## 7 · Alternating series

For Σ (−1)n bₙ with bₙ ≥ 0: if bₙ decreases monotonically to 0, the series **converges** — cancellation wins even when Σ|aₙ| diverges. Bonus: the error from truncating after N terms is smaller than the first omitted term, bN+1. When absolute convergence fails but the alternating test succeeds, the convergence is called **conditional**.

**Worked example**

**Example 1 — Sum a geometric series.** Σn=0∞ 4·(2/3)ⁿ.

Step 1: Identify a = 4 (the n = 0 term) and r = 2/3.

Step 2: Check |r| = 2/3 < 1 — geometric theory applies.

Step 3: Sum = a/(1 − r) = 4/(1 − 2/3) = 4/(1/3) = **12**.

**Example 2 — Two famous limits, numerically.**

(1 + 1/n)ⁿ: n = 10 → 2.594, n = 1000 → 2.717 — creeping toward e.

n1/n: n = 10 → 1.259, n = 1000 → ≈ 1.007 — settling toward 1.

**Example 3 — A series doomed by the nth-term test.** Σ n/(2n + 3): the terms approach ½, never 0, so the series diverges — no further work needed.

**Watch out**

Never run the geometric formula while quoting |r| < 1 without checking it: r = 3/2 gives the nonsense 1/(1 − 3/2) = −2 for a visibly positive, exploding series. Also remember the nth-term test is one-directional — aₙ → 0 is necessary, never sufficient.

**Chapter formula sheet**

lim a/(a + n) = 1 · lim n/(2n + 7) = 1/2 (divide by highest power)

lim (1 + 1/n)ⁿ = e ≈ 2.71828 · lim n1/n = 1

Σn=0∞ arⁿ = a/(1 − r) ⇔ |r| < 1 · Σn=k∞ arⁿ = ark/(1 − r)

nth-term test: aₙ ↛ 0 ⇒ Σaₙ diverges

p-test: Σ 1/np converges ⇔ p > 1

Ratio L = lim |aₙ₊₁/aₙ| · Root L = lim |aₙ|1/n — L < 1 converge, L > 1 diverge, L = 1 silent

Integral test: Σ f(n) and ∫1∞ f(x) dx share fate (f positive, decreasing)

Alternating: bₙ ↘ 0 ⇒ converges · error ≤ first omitted term

## Common mistakes

- Reading aₙ → 0 as proof of convergence — the harmonic series Σ 1/n diverges.

- Plugging |r| ≥ 1 into the geometric sum formula.

- Applying the ratio or root test and declaring a verdict when L = 1 — it is inconclusive.

- Using the integral test on a function that is not eventually decreasing.

- Treating Σ (−1)ⁿ/n as absolutely convergent — it is conditional.

- Computing the first few partial sums and guessing the series settles — decide with tests, not vibes.

## Practice exercises

1. The sequence aₙ = n / (n + 5) converges to:

- A. 1
- B. 0
- C. 5
- D. the sequence diverges

2. Compute Σ n=0 ∞ (1/3)ⁿ.

3. Classify Σ n=1 ∞ n / (2n + 7) — converge or diverge?

4. The series Σ n=1 ∞ 1 / n 3/2 is:

- A. convergent (p = 3/2 > 1)
- B. divergent (it is a p-series)
- C. divergent, since the terms approach 0 slowly
- D. convergent only if paired with an alternating sign

5. What does (1 + 1/n)ⁿ approach as n → ∞? Give the decimal to two decimals.

6. Σ n=1 ∞ (−1)ⁿ / n² converges — absolutely or conditionally?

7. Apply the ratio test to Σ n=0 ∞ 2ⁿ / n! . The limit L = lim |aₙ₊₁/aₙ| is:

- A. 0, so the series converges
- B. 2, so the series diverges
- C. 1, so the test is inconclusive
- D. ∞, so the series diverges

8. Compute Σ n=1 ∞ (1/4)ⁿ.

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