What is an infinite series?
A sequence is a list of numbers: a₁, a₂, a₃, … A series is what you get when you add them: a₁ + a₂ + a₃ + … If the adding never stops, it is an infinite series, written Σ aₙ (n from 1 to ∞).
We cannot add forever, so we look at partial sums:
S₁ = a₁, S₂ = a₁ + a₂, S₃ = a₁ + a₂ + a₃, …, Sₙ = a₁ + … + aₙ.
The partial sums form a new sequence. Its behaviour decides everything.
Convergent and divergent series
If Sₙ gets as close as we like to one number S as n grows (lim Sₙ = S), the series converges and its sum is S.
If Sₙ grows without limit, or keeps jumping without settling, the series diverges. Examples: 1 + 1 + 1 + … (grows), 1 − 1 + 1 − 1 + … (jumps between 1 and 0).
nth-term test for divergence: if aₙ does not go to 0, the series diverges. But if aₙ → 0, we still need another test.
Geometric series and their uses
A geometric series has a fixed ratio r between terms: a + ar + ar² + ar³ + …
Partial sum: Sₙ = a(1 − rⁿ) ÷ (1 − r), for r ≠ 1.
If |r| < 1, then rⁿ → 0, so S = a ÷ (1 − r). If |r| ≥ 1 (and a ≠ 0), the series diverges.
Uses:
- Recurring decimals: 0.444… = 0.4 + 0.04 + … = 0.4 ÷ 0.9 = 4/9.
- Bouncing balls: total distance travelled.
- Medicine doses: the amount of a drug in the body after many equal daily doses levels off.
- Money: the present value of a payment that goes on forever.
- Zeno's paradox: infinitely many smaller and smaller steps can take a finite time.
Harmonic series and p-series
The harmonic series 1 + 1/2 + 1/3 + 1/4 + … diverges. Group the terms: 1/3 + 1/4 > 1/2, then 1/5 + … + 1/8 > 1/2, and so on. We keep adding at least 1/2 again and again, so the sum has no limit (it grows like ln n, very slowly).
A p-series is Σ 1/nᵖ.
- p > 1: converges (for example Σ 1/n² = π²/6 ≈ 1.645).
- p ≤ 1: diverges (p = 1 is the harmonic series).
Tests for convergence
For series with positive terms (except the alternating test):
- Integral test: if f(n) = aₙ with f positive, continuous and decreasing, then Σ aₙ and ∫₁^∞ f(x) dx both converge or both diverge.
- Direct comparison: if 0 ≤ aₙ ≤ bₙ and Σ bₙ converges, so does Σ aₙ. If aₙ ≥ bₙ ≥ 0 and Σ bₙ diverges, so does Σ aₙ.
- Limit comparison: if aₙ/bₙ → a positive finite number, both series behave the same.
- Ratio test: let L = lim |aₙ₊₁ / aₙ|. L < 1 converges (absolutely), L > 1 diverges, L = 1 tells nothing.
- Alternating series test: Σ (−1)ⁿ⁺¹ bₙ with bₙ positive, decreasing and → 0 converges. The error after n terms is at most the next term bₙ₊₁.
Absolute vs conditional: if Σ |aₙ| converges, the series converges absolutely. If Σ aₙ converges but Σ |aₙ| does not, it converges conditionally. Example: 1 − 1/2 + 1/3 − … converges (to ln 2) but only conditionally.
Looking ahead: power series and Taylor series
A power series Σ cₙ(x − a)ⁿ is an infinite series with x inside. The ratio test gives its radius of convergence R: it converges for |x − a| < R; the end points are checked separately.
Example: 1 + x + x² + … = 1/(1 − x) for |x| < 1 (a geometric series with r = x).
Taylor series write a function as a power series: f(x) = Σ f⁽ⁿ⁾(a)(x − a)ⁿ / n!. With a = 0 it is a Maclaurin series, e.g. eˣ = 1 + x + x²/2! + x³/3! + … Stopping after a few terms gives a Taylor polynomial; the Lagrange error bound says how far off it can be.
Try it: tear a paper strip
Take a strip of paper 20 cm long. Tear off half (10 cm) and put it on the table. Tear half of what is left (5 cm) and place it next to the first. Keep going. Your line on the table creeps towards 20 cm but never passes it. That is 10 + 5 + 2.5 + … = 10 ÷ (1 − 1/2) = 20. Then try predicting with the a and r sliders in the last 3D step.
Key formulas and definitions
- Sₙ = a₁ + a₂ + … + aₙ; series converges if lim Sₙ = S exists
- Geometric: Sₙ = a(1 − rⁿ)/(1 − r); S∞ = a/(1 − r) for |r| < 1
- nth-term test: aₙ ↛ 0 ⇒ diverges
- p-series Σ 1/nᵖ: converges if p > 1, diverges if p ≤ 1
- Ratio test: L = lim |aₙ₊₁/aₙ|; L < 1 converges, L > 1 diverges, L = 1 inconclusive
- Alternating series: |S − Sₙ| ≤ bₙ₊₁
Worked examples
1. Find the sum of 6 + 3 + 1.5 + 0.75 + …
a = 6, r = 3/6 = 0.5. |r| < 1, so S = a/(1 − r) = 6/0.5 = 12.
2. Write 0.272727… as a fraction.
0.27 + 0.0027 + … is geometric with a = 0.27, r = 0.01. S = 0.27/0.99 = 27/99 = 3/11.
3. Does Σ n/(n + 1) converge?
aₙ = n/(n + 1) → 1, not 0. By the nth-term test the series diverges.
4. A ball is dropped from 2 m and each bounce reaches half the previous height. Find the total distance it travels.
Down 2 m, then up and down 1 m, 0.5 m, … Total = 2 + 2(1 + 0.5 + 0.25 + …) = 2 + 2 × 1/(1 − 0.5) = 2 + 4 = 6 m.
5. Does Σ 1/(n² + 3) converge?
0 < 1/(n² + 3) < 1/n². Σ 1/n² is a p-series with p = 2 > 1, so it converges. By direct comparison, Σ 1/(n² + 3) converges.
6. Use the ratio test on Σ 3ⁿ/n!.
aₙ₊₁/aₙ = 3ⁿ⁺¹/(n + 1)! × n!/3ⁿ = 3/(n + 1) → 0. L = 0 < 1, so the series converges (absolutely).
7. Does 1 − 1/√2 + 1/√3 − 1/√4 + … converge absolutely, conditionally or not at all?
bₙ = 1/√n is positive, decreasing and → 0, so the alternating series converges. But Σ 1/√n is a p-series with p = 1/2 ≤ 1, which diverges. So it converges conditionally.
8. For which x does Σ (x/3)ⁿ converge, and to what?
Geometric with r = x/3. It converges when |x/3| < 1, i.e. −3 < x < 3. Sum = 1/(1 − x/3) = 3/(3 − x) (starting from n = 0).
Common mistakes
- Thinking aₙ → 0 proves convergence. It is only necessary; the harmonic series has aₙ → 0 but diverges.
- Using a/(1 − r) when |r| ≥ 1. The formula only works for −1 < r < 1.
- Mixing up the sequence and the series: aₙ → 0 is about terms; convergence of a series is about partial sums Sₙ.
- Saying the ratio test proves divergence when L = 1. L = 1 gives no answer; use another test.