Sequences and Series: Basic Concepts

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.
• n^1/n → 1. n^1/n = e^{ln n / n} and ln n / n → 0, so 1000^1/1000 ≈ 1.007.

4 · From sequences to series; the geometric engine

A series Σ aₙ asks whether the running totals S_N = 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 → ∞, r^N+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: Σ_n=k^∞ arⁿ = ar^k/(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/n^p converges iff p > 1. (The integral test proves it: ∫₁^∞ dx/x^p and Σ 1/n^p share their fate exactly when f(x) = 1/x^p 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 Σ_n>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, b_N+1. When absolute convergence fails but the alternating test succeeds, the convergence is called conditional.

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.