# Chapter 23: How to Solve Series Problems

> A mechanical toolkit: nth-term first, then geometric, ratio, root, comparison and integral tests — plus a flowchart for any series you meet.

**Canonical:** https://calculus.kitzo.work/chapter/series-solving  
**Level:** Advanced  
**Estimated time:** 45 minutes  
**Exercises:** 8 (auto-marked; answers stay server-side)

## Topics
- Geometric sums
- Ratio test in practice
- Root test in practice
- Integral test in practice
- Choosing comparison tests
- Alternating & mixed-sign series
- The series strategy flowchart

**Why this matters**

Series problems are pattern recognition under pressure. Almost every exam series falls to one of six moves, and the trick is knowing which to reach for first — a fixed order that turns panic into procedure.

## 1 · Geometric sums, mechanically

The geometric series is the only everyday series with a closed-form sum: `Σn=0∞ arn = a/(1 − r)`, valid **iff |r| < 1**. Mechanics to drill:

- **Find a**: the value of the FIRST term actually included — not always the n = 0 term. Σn=k∞ arn = ark/(1 − r).

- **Find r**: divide any term by its predecessor.

- **|r| ≥ 1 ⇒ diverges**. r = 2 or r = −1 never sums. Say it out loud every time.

## 2 · The nth-term test goes first

Before any machinery, compute lim aₙ. If it is not zero, the series diverges — done in ten seconds. Terms growing like (3n + 1)/(n + 2) → 3 can never add to a finite number. This check costs nothing and kills a fifth of exam problems outright.

## 3 · The ratio test in practice

Let `L = lim |aₙ₊₁ / aₙ|`. Then L < 1 ⇒ converges absolutely; L > 1 or L = ∞ ⇒ diverges; L = 1 ⇒ silent. Two battlefields:

- **Factorials**: (n + 1)!/n! = n + 1 — factorial ratios simplify brutally. Σ 3ⁿ/n!: ratio = 3/(n + 1) → 0, converges.

- **Exponentials**: an+1/an = a. Σ n²/2ⁿ: ratio → ½, converges.

Work slowly: write aₙ₊₁, write the quotient, cancel, then take the limit. Never simplify inside the absolute values prematurely.

## 4 · The root test in practice

Let `L = lim |aₙ|1/n`, same trichotomy. Reach for it when the whole term is trapped inside an nth power: `( (2n + 1)/(3n + 2) )ⁿ` gives L = 2/3 < 1, converges. Uses in disguise: (n!)1/n ~ n/e via Stirling — handy only with that fact in pocket, otherwise prefer the ratio test on factorials.

## 5 · Choosing between comparison tests

Against a positive-term series, ask: *what does it behave like at ∞?*

- **p-test benchmark**: keep only leading powers → 1/np. Converges iff p > 1. Σ (n³ + 1)/(n⁵ + n) ~ Σ 1/n² ✓ converges.

- **Limit comparison** (ratio of leading terms, L ∈ (0, ∞)): the default for rational/algebraic terms.

- **Direct comparison**: when bounding is easy and the inequality is clean — e.g. sin⁴n / n² ≤ 1/n².

- **Integral test**: when f(x) is smooth, positive, decreasing, and integrable. Σ 1/(n ln n): ∫2∞ dx/(x ln x) = ln ln x → ∞, so the series diverges.

## 6 · The integral test in practice

Recipe: define f(x) with f(n) = aₙ; check positive for x ≥ N; check decreasing (derivative negative, or ratio aₙ₊₁/aₙ ≤ 1); integrate from N to ∞. Bonus gift — the remainder satisfies `Σn>N f(n) ≤ ∫N∞ f(x) dx`, so the integral also estimates truncation error, not just fate.

## 7 · Mixed signs and the grand flowchart

For non-positive series: first test Σ |aₙ| (everything above applies). If it converges, you are done — absolute convergence. If not, try the **alternating series test**: terms alternate and decrease in magnitude to 0 ⇒ converges (conditionally). Otherwise it likely diverges — confirm with the nth-term test.

The universal flowchart: **(1)** nth-term test → kill if aₙ ↛ 0. **(2)** Geometric? → closed form. **(3)** Factorials or aⁿ? → ratio. **(4)** nth power everywhere? → root. **(5)** Algebraic/rational? → limit comparison with a p-series. **(6)** Smooth decreasing? → integral test. **(7)** Alternating/mixed signs? → absolute first, then alternating test.

**Worked example**

**Example 1 — Geometric, starting at n = 2.** Σn=2∞ 5/3ⁿ.

Step 1: First included term is 5/9; ratio r = 1/3.

Step 2: |r| < 1 ✓. Sum = (5/9)/(1 − 1/3) = (5/9)·(3/2) = **5/6**.

**Example 2 — Ratio test on Σ 3ⁿ/n!.**

Step 1: aₙ₊₁/aₙ = 3ⁿ⁺¹/(n+1)! · n!/3ⁿ = 3/(n + 1).

Step 2: L = lim 3/(n + 1) = 0 < 1 ⇒ converges absolutely.

**Example 3 — Integral test on Σ 1/(n ln n), n ≥ 2.**

Step 1: f(x) = 1/(x ln x) > 0, decreasing for x ≥ 2.

Step 2: ∫2∞ dx/(x ln x) = [ln ln x]2∞ = ∞.

Step 3: The integral diverges, so the series **diverges**.

**Watch out**

L = 1 is a shrug, not a verdict. The harmonic series (L = 1, diverges) and Σ 1/n² (L = 1, converges) both sit there — fall back to comparison or the integral test when the ratio test goes quiet.

**Chapter formula sheet**

Σn=k∞ arⁿ = ark/(1 − r) ⇔ |r| < 1

nth-term test: aₙ ↛ 0 ⇒ diverge · factorial shortcut (n+1)!/n! = n+1

Ratio L = lim |aₙ₊₁/aₙ|: L < 1 conv · L > 1 div · L = 1 silent

Root L = lim |aₙ|1/n: same trichotomy

Limit comparison: lim aₙ/bₙ = L ∈ (0, ∞) ⇒ same fate · benchmark Σ 1/np

Integral test: Σ f(n) vs ∫1∞ f(x) dx — both positive, f decreasing

Remainder after N: Σn>N aₙ ≤ ∫N∞ f(x) dx

Absolute conv ⇒ conv · Alternating bₙ ↘ 0 ⇒ conv, error ≤ first omitted term

## Common mistakes

- Quoting the geometric sum with |r| ≥ 1 — check |r| first, always.

- Skipping the nth-term test and reaching for the ratio test on a doomed series.

- Deciding on L = 1 as if the test had spoken — it is inconclusive.

- Picking the wrong comparison partner (1/x vs 1/x²); match degrees of growth, not vibes.

- Applying the integral test to a non-decreasing f.

- Declaring Σ (−1)ⁿ⁺¹/n absolute — it is only conditional; Σ 1/n diverges.

## Practice exercises

1. The sum of Σ n=0 ∞ 2·(1/4)ⁿ is:

- A. 8/3
- B. 2
- C. 4/3
- D. the series diverges

2. Compute Σ n=1 ∞ (1/2)ⁿ.

3. Apply the ratio test to Σ n=1 ∞ n! / 2ⁿ. The conclusion is:

- A. diverges — L = lim (n + 1)/2 = ∞ > 1
- B. converges — L = 1/2 < 1
- C. inconclusive — L = 1
- D. converges — factorials dominate exponentials

4. Compute Σ n=0 ∞ (3/10)ⁿ. Give the decimal to two decimals.

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

6. Which of these converges?

- A. Σ 1/n
- B. Σ 1/n²
- C. Σ n/(n + 1)
- D. Σ 3 (constant terms)

7. In the integral-test argument for Σ 1/n³, evaluate ∫ 1 ∞ dx / x³.

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

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