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Part IA Michaelmas, Lent, Easter Term

Series Convergence

Definitions

A series is ∑k=0∞uk\sum_{k=0}^{\infty} u_k. The partial sums are Sn=∑k=0nukS_n = \sum_{k=0}^{n} u_k.

Convergence: The series converges if lim⁡n→∞Sn=S\lim_{n \to \infty} S_n = S exists (finite). We write ∑uk=S\sum u_k = S.

Formal definition: For any ε>0\varepsilon > 0, there exists NN such that ∣S−Sn∣<ε|S - S_n| < \varepsilon for all n>Nn > N.


Necessary Condition for Convergence

If ∑uk\sum u_k converges, then lim⁡k→∞uk=0\lim_{k \to \infty} u_k = 0.

Warning: The converse is false. Example: harmonic series ∑1k\sum \frac{1}{k} diverges even though 1k→0\frac{1}{k} \to 0.


Absolute vs Conditional Convergence

Absolute convergence: ∑∣uk∣\sum |u_k| converges.

Conditional convergence: ∑uk\sum u_k converges but ∑∣uk∣\sum |u_k| diverges.

Key theorem: If ∑∣uk∣\sum |u_k| converges, then ∑uk\sum u_k converges.

Riemann rearrangement theorem: Conditionally convergent series can be rearranged to converge to any value (or diverge). Absolutely convergent series are immune to rearrangement.


Comparison Test

For series with 0≤uk≤vk0 \leq u_k \leq v_k:

  1. If ∑vk\sum v_k converges, then ∑uk\sum u_k converges
  2. If ∑uk\sum u_k diverges, then ∑vk\sum v_k diverges

Limit comparison: If lim⁡k→∞ukvk=L\lim_{k \to \infty} \frac{u_k}{v_k} = L with 0<L<∞0 < L < \infty, then ∑uk\sum u_k and ∑vk\sum v_k behave the same.

Example: ∑1k(k+1)\sum \frac{1}{k(k+1)} converges since 1k(k+1)<1k2\frac{1}{k(k+1)} < \frac{1}{k^2} and ∑1k2\sum \frac{1}{k^2} converges.


Ratio Test

Let L=lim⁡k→∞∣uk+1uk∣L = \lim_{k \to \infty} \left|\frac{u_{k+1}}{u_k}\right|:

  • L<1L < 1: series converges absolutely
  • L>1L > 1: series diverges
  • L=1L = 1: test is inconclusive

Example: ∑k22k\sum \frac{k^2}{2^k}. Ratio: (k+1)2/2k+1k2/2k=12(1+1k)2→12<1\frac{(k+1)^2/2^{k+1}}{k^2/2^k} = \frac{1}{2}\left(1 + \frac{1}{k}\right)^2 \to \frac{1}{2} < 1. Converges.


Harmonic Series Divergence

∑k=1∞1k=1+12+13+14+⋯\sum_{k=1}^{\infty} \frac{1}{k} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots

Harmonic series partial sums, showing slow divergence with proof by grouping terms into blocks > 1/2 each

Proof of divergence (grouping):

1+12+(13+14)+(15+16+17+18)+⋯1 + \frac{1}{2} + \left(\frac{1}{3} + \frac{1}{4}\right) + \left(\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8}\right) + \cdots

>1+12+12+12+⋯> 1 + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \cdots

The partial sums grow without bound.


Alternating Series (Leibniz Criterion)

An alternating series ∑(−1)kak\sum (-1)^k a_k with ak>0a_k > 0 converges if:

  1. aka_k is monotonically decreasing for large kk
  2. lim⁡k→∞ak=0\lim_{k \to \infty} a_k = 0

Example: The alternating harmonic series: ∑k=1∞(−1)k+1k=1−12+13−14+⋯=ln⁡2\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k} = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots = \ln 2

This is conditionally convergent (not absolutely, since harmonic series diverges).


Riemann Zeta Function

ζ(p)=∑k=1∞1kp\zeta(p) = \sum_{k=1}^{\infty} \frac{1}{k^p}

  • Converges for p>1p > 1
  • Diverges for p≤1p \leq 1

Special values:

  • ζ(2)=π26\zeta(2) = \frac{\pi^2}{6}
  • ζ(4)=π490\zeta(4) = \frac{\pi^4}{90}

Radius of Convergence for Power Series

For ∑anxn\sum a_n x^n, the radius of convergence is:

R=lim⁡n→∞∣anan+1∣(if limit exists)R = \lim_{n \to \infty} \left|\frac{a_n}{a_{n+1}}\right| \quad \text{(if limit exists)}

  • Converges absolutely for ∣x∣<R|x| < R
  • Diverges for ∣x∣>R|x| > R
  • At x=±Rx = \pm R: must check separately

Example: ∑xnn\sum \frac{x^n}{n}.

R=lim⁡1/n1/(n+1)=1R = \lim \frac{1/n}{1/(n+1)} = 1. Converges for ∣x∣<1|x| < 1.

At x=1x = 1: harmonic series diverges. At x=−1x = -1: alternating harmonic converges.


Summary: Choosing a Test

SituationPreferred Test
Comparison with known seriesComparison test
Factorials, exponentialsRatio test
Alternating signsLeibniz criterion
Powers 1/kp1/k^pp-test (zeta)
Power seriesRadius of convergence

Complexity: Ratio test is often easiest to apply. Comparison test requires knowing a suitable reference series.