MAST20029 · Engineering Mathematics
Sequences and Series: the Convergence Toolkit
This chapter is a toolkit for deciding whether a series converges and, for power series, where. You run the divergence test first, then choose comparison, limit-comparison, ratio, root, integral or Leibniz by the SHAPE of the terms, and distinguish absolute from conditional convergence. Power series give a radius and interval (test the endpoints), and Taylor polynomials/series with the Lagrange remainder let you approximate functions and even sum numerical series in closed form. Examined as Q7–Q8.
What this chapter covers
- 011. Partial sums Sₙ and the meaning of Σ aₙ = lim Sₙ
- 022. Geometric series Σ a rⁿ = a/(1−r) for |r|<1; p-series Σ 1/nᵖ (converges iff p>1)
- 033. Divergence (nth-term) test: if aₙ does not → 0, the series diverges (do first)
- 044. Comparison and limit-comparison tests for non-negative terms
- 055. Ratio and root tests (L<1 converge, L>1 diverge, L=1 inconclusive)
- 066. Integral test (positive, continuous, decreasing) and Leibniz alternating test
- 077. Absolute vs conditional convergence; standard limits like (1+1/n)ⁿ → e
- 088. Power series radius/interval; Taylor polynomials, series and the remainder bound
Comparison test and radius of convergence
- 1 mark(a) Bound the terms: since −1 ≤ sin n ≤ 1, we have 0 ≤ (2 + sin n)/(n² + 1) ≤ 3/n². The numerator is bounded; do not try to evaluate sin n.
- 1 markCompare with Σ 3/n², a convergent p-series (p = 2 > 1). By the comparison test the original series CONVERGES.
- 1 mark(b) Apply the ratio test to |aₙ|: |a_(n+1)/aₙ| = ((n+1)!/n!) · (nⁿ/(n+1)^(n+1)) |x − 3| = nⁿ/(n+1)ⁿ |x − 3| = (n/(n+1))ⁿ |x − 3|.
- 1 markTake the limit: (n/(n+1))ⁿ = (1 + 1/n)^(−n) → e^(−1), so L = |x − 3|/e (using the provided standard limit (1+1/n)ⁿ → e).
- 2 marksConverge when L < 1: |x − 3| < e, so the radius of convergence is R = e.
Key terms
- Partial sum
- Sₙ = a₁ + a₂ + ... + aₙ; the series Σ aₙ converges to S if Sₙ → S. Geometric and telescoping series have closed-form partial sums.
- Geometric / p-series
- Σ a rⁿ converges to a/(1−r) iff |r| < 1; the p-series Σ 1/nᵖ converges iff p > 1 (the harmonic series p = 1 diverges). Both are standard comparison benchmarks.
- Divergence test
- If aₙ does not tend to 0, then Σ aₙ diverges. It is one-directional (aₙ → 0 does NOT prove convergence) and should be your first check.
- Ratio / root test
- Compute L = lim |a_(n+1)/aₙ| (ratio) or L = lim |aₙ|^(1/n) (root): L < 1 converges, L > 1 diverges, L = 1 is inconclusive. Best for factorials and nth powers.
- Absolute vs conditional
- A series converges absolutely if Σ |aₙ| converges; if Σ aₙ converges but Σ |aₙ| does not, convergence is conditional (typical of alternating series passing only Leibniz).
- Radius of convergence
- For a power series Σ cₙ(x−a)ⁿ, the value R such that the series converges for |x−a| < R; found by the ratio or root test, with the endpoints tested separately.
Sequences and Series: the Convergence Toolkit FAQ
Which convergence test should I use?
Always run the divergence test first — if the terms do not go to zero, you are done. Then choose by the shape of aₙ: factorials or nth powers suggest the ratio or root test; a rational or bounded-over-power form suggests comparison or limit comparison with a p-series; a positive decreasing term you can integrate suggests the integral test; an alternating sign suggests Leibniz. Match the tool to the structure.
What is the difference between absolute and conditional convergence?
A series converges absolutely if the series of absolute values Σ |aₙ| converges; then the original converges too. If Σ aₙ converges but Σ |aₙ| diverges, the convergence is conditional — typical of an alternating series that passes the Leibniz test but whose absolute version (like the harmonic series) diverges.
How do I find the radius and interval of convergence?
Apply the ratio (or root) test to |cₙ(x−a)ⁿ|, set the limit L < 1, and solve for |x−a| < R to read off the radius R. Then test the two endpoints x = a − R and x = a + R separately with another test, since the ratio test is inconclusive there. The interval may or may not include each endpoint.
How are Taylor series used to sum a numerical series?
Recognise that a numerical series is a known Taylor series evaluated at a specific point. For example, since e^(cx) = Σ (cx)ⁿ/n!, setting x = 1 gives Σ cⁿ/n! = e^c. Build the Taylor series, then substitute the value that turns it into your target sum — a recurring Q8 trick.
Exam move
Build a decision tree and practise it until choosing a test is instant: divergence test first, then read the shape of aₙ to pick comparison/limit-comparison (rational, bounded), ratio/root (factorials, nth powers), integral (positive decreasing), or Leibniz (alternating). Memorise the benchmarks — geometric (|r|<1) and p-series (p>1) — and the provided standard limits, especially (1+1/n)ⁿ → e, which appears in radius problems. For power series, always test the endpoints separately after finding R. For Taylor questions, learn the standard series and the trick of evaluating them at a point to sum a numerical series, and quote the test or series you used.
