Chapter 23: How to Solve Series Problems

Why this mattersSeries 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: Σ_n=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. Σ_n=k^∞ arⁿ = ar^k/(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/n^p. 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 Σ_n>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. Σ_n=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 outL = 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 Σ_n=k^∞ arⁿ = ar^k/(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/n^p
Integral test: Σ f(n) vs ∫₁^∞ f(x) dx — both positive, f decreasing
Remainder after N: Σ_n>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.