Lesson 15: Sequences and Series: Convergence Tests

A series is a sum with infinitely many terms — and its meaning is a limit: the series Σ a_n converges to S when the sequence of partial sums S_N = a_1 + ... + a_N approaches S. This lesson is about deciding convergence (often without finding the sum), the handful of series whose sums we can find exactly, and the tests that classify the rest. The mindset shift from earlier chapters: here the question is usually not "what is it worth?" but "does it settle at all?" — and that question turns out to be the gateway to Lesson 16, where series become functions.

The series we can sum: geometric and telescoping

Geometric: Σ from n=0 of r^n = 1/(1 − r) for |r| < 1, divergent for |r| ≥ 1. So Σ (1/3)^n = 1/(1 − 1/3) = 3/2, and Σ from n=1 of (2/5)^n = (2/5)/(1 − 2/5) = 2/3. The first term and the ratio are the whole story. Telescoping: Σ 1/(n(n + 1)): partial fractions (Lesson 13!) give 1/n − 1/(n + 1), consecutive terms cancel, and the partial sum is 1 − 1/(N + 1) → 1. Two families, two tricks: recognise a constant ratio, or engineer a cancellation.

The first test costs nothing: divergence

If a_n does not tend to 0, the series diverges — full stop. Σ n/(n + 1) diverges because its terms tend to 1. Σ (−1)^n diverges (terms never settle). But the converse is false, and the counterexample runs the world: the harmonic series Σ 1/n has terms tending to 0 yet diverges (grouping: 1/3 + 1/4 > 1/2, 1/5 + ... + 1/8 > 1/2, forever). Terms → 0 is the entry fee, not the prize.

The standard tests

• p-series: Σ 1/n^p converges iff p > 1 (the improper-integral p-test of Lesson 14, inherited via the integral test). Σ 1/n^2 converges; Σ 1/sqrt(n) diverges.
• Comparison: compare term-by-term with a known series. Σ 1/(n^2 + 3) < Σ 1/n^2 → converges. Σ 1/(n − 1)... for n ≥ 2 terms exceed 1/n eventually → diverges by comparison with harmonic (limit comparison handles the bookkeeping).
• Ratio test: L = lim |a_(n+1)/a_n|. L < 1: converges; L > 1: diverges; L = 1: no information (harmonic and Σ 1/n^2 both give L = 1). Factorials and exponentials are its home turf: Σ n!/10^n has ratio (n + 1)/10 → ∞, diverges (factorial outgrows exponential — the growth hierarchy again); Σ 2^n/n! has ratio 2/(n + 1) → 0, converges.
• Alternating series: Σ (−1)^n b_n with b_n decreasing to 0 converges. Σ (−1)^n/n converges — but Σ 1/n diverges, so this convergence is conditional, not absolute; the distinction matters in Lesson 16.

Worked Example 1 (classification drill). (i) Σ 3/(n^2 + n): terms ~ 3/n^2, converges by limit comparison with Σ 1/n^2. (ii) Σ n^2/2^n: ratio (n + 1)^2/(2 n^2) → 1/2 < 1, converges. (iii) Σ (n + 1)/(2n + 1): terms → 1/2 ≠ 0, diverges by the divergence test. Three series, three different tests, no sums computed — this is the daily work of the subject.

Application: dosing and bouncing — geometric series in disguise

Medication. A patient takes 100 mg daily; each day the body eliminates 40% of whatever is present, so 60% carries over. Total in the body just after the n-th dose: 100 + 100(0.6) + 100(0.6)^2 + ... — a geometric series with limit 100/(1 − 0.6) = 250 mg. The plateau is the infinite sum; pharmacology runs on it. Bouncing ball. Dropped from 10 m, rebounding to 0.6 of each height: total distance = 10 + 2·10·(0.6 + 0.6^2 + ...) = 10 + 20·(0.6/0.4) = 10 + 30 = 40 m. (The first drop is counted once, every later height twice — up and down.) The ball bounces infinitely often in finite total distance and finite total time, a friendly Zeno paradox with a convergent answer.

Common mistakes

• "Terms → 0, so it converges." The harmonic series refutes this permanently. The divergence test has one direction only: terms not → 0 proves divergence; terms → 0 proves nothing.
• Ratio test L = 1 declared as convergence (or divergence). L = 1 is silence. Both Σ 1/n (divergent) and Σ 1/n^2 (convergent) produce L = 1 — switch tests.
• Geometric formula with |r| ≥ 1. Σ 2^n is not 1/(1 − 2) = −1; a sum of positive terms cannot be negative. Check |r| < 1 before any geometric formula — the formula's derivation assumes it.
• Confusing the sequence with the series. a_n → 3 means the sequence converges and the series Σ a_n diverges (terms do not → 0). The words "it converges" are meaningless until you say what "it" is.

Exam Experience

1. Σ from n=1 of (3/7)^n equals

• (A) 3/4
• (B) 7/4
• (C) 4/3
• (D) 3/7

Answer: (A). Geometric with first term 3/7 and ratio 3/7: sum = (3/7)/(1 − 3/7) = (3/7)/(4/7) = 3/4.

2. Which series converges?

• (A) Σ n/(n + 5)
• (B) Σ 1/n^(1/2)
• (C) Σ 1/n^2
• (D) Σ (−2)^n

Answer: (C). p-series with p = 2 > 1. (A) fails the divergence test (terms → 1); (B) is p = 1/2; (D) is geometric with |r| = 2 ≥ 1.

3. The ratio test applied to Σ 5^n/n! gives L =

• (A) 0
• (B) 1
• (C) 5
• (D) ∞

Answer: (A). a_(n+1)/a_n = 5/(n + 1) → 0 < 1, so the series converges — factorials outgrow exponentials.

Free-response question. (a) Determine whether Σ from n=1 of n/(n^2 + 1) converges or diverges, with justification. (b) Find the exact sum of Σ from n=0 of ( (1/4)^n + (1/3)^n ). (c) A drug is dosed at 50 mg every 12 hours; 30% of the amount present is eliminated between doses. Find the plateau amount in the body just after a dose, in the long run.

Scoring guideline.

• (a) 1 point. Diverges by limit comparison with Σ 1/n: the ratio of terms (n/(n^2 + 1))/(1/n) = n^2/(n^2 + 1) → 1 > 0, and the harmonic series diverges. (Comparison n/(n^2 + 1) > 1/(2n) for n ≥ 1 also earns it.)
• (b) 1 point. Sum of two geometric series: 1/(1 − 1/4) + 1/(1 − 1/3) = 4/3 + 3/2 = 17/6.
• (c) 1 point. Carry-over ratio 0.7; plateau = 50/(1 − 0.7) = 500/3 ≈ 166.7 mg. The geometric-series set-up must be visible (50 + 50(0.7) + ...), not just the final number.

Going deeper: the integral test, Basel, and alternating estimates

The integral test binds series to Lesson 14: for positive decreasing f, Σ f(n) and ∫ f(x) dx converge or diverge together. It also bounds sums. For Σ 1/n^2: the tail beyond n = 1 lies between ∫ from 2 to ∞ of dx/x^2 = 1/2 and ∫ from 1 to ∞ of dx/x^2 = 1, so the total sits between 1.5 and 2. Euler famously closed the question in 1734: Σ 1/n^2 = π^2/6 ≈ 1.645 — π materialising inside a sum of rational squares, still startling after three centuries. (His proof, via the sine series of Lesson 16, is a fine piece of enrichment reading.) Alternating estimates. The alternating harmonic series sums to ln 2 ≈ 0.6931, and the alternating-series estimation theorem says the error after N terms is at most the first omitted term. After 10 terms the partial sum is 0.6456; the error is 0.0475, indeed below 1/11 ≈ 0.091 ✓. Slow but certified — contrast the Leibniz series of Lesson 16, whose error shrinks like 1/(2N) and needs thousands of terms for three decimals. When you estimate with series, the error theorem is the product; the digits are packaging. Finally, absolute convergence: Σ (−1)^n/n^2 converges absolutely (since Σ 1/n^2 converges) and is therefore robust — rearrangements cannot change its sum. The alternating harmonic series converges only conditionally, and a theorem of Riemann says its terms can be rearranged to sum to any prescribed value, or to diverge. Conditional convergence is a genuine, delicate phenomenon — the point where infinite addition stops behaving like finite addition, and the reason this course insists on the distinction.

Going deeper: the root test, and two calibrations

The root test reads the geometric skeleton directly: for Σ a_n, examine lim |a_n|^(1/n) = r; convergent if r < 1, divergent if r > 1, silent at r = 1. It shines where the ratio test is clumsy — powers of n as exponents. Σ (n/(2n + 1))^n: the n-th root is n/(2n + 1) → 1/2 < 1, convergent. Σ (2n/(2n + 1))^n would give root → 1... instead take Σ n^2/3^n under the ratio test: a_(n+1)/a_n = ((n + 1)/n)^2 · (1/3) → 1/3 < 1, convergent — polynomial growth divided by geometric growth always loses, which is the growth ladder of Lesson 12 in series clothing. And the cautionary calibration: for Σ 1/n the root tends to 1 and the ratio tends to 1 — both tests fall silent exactly on the boundary series of Lesson 15's main text. Boundary cases are not failures of diligence; they are where mathematics keeps its subtleties, and the comparison family exists precisely for them.

One more telescoping calibration, with a gap: Σ 1/(n(n + 2)). Partial fractions give 1/(n(n + 2)) = (1/2)(1/n − 1/(n + 2)), and because the offset is 2, two terms survive at each end of the partial sum: S_N = (1/2)(1 + 1/2 − 1/(N + 1) − 1/(N + 2)) → (1/2)(3/2) = 3/4. The cancellation pattern — write out the first four and last four terms and watch the middle die — is the reliable way to handle any offset; formulae memorised for offset 1 misfire silently on offset 2.