Skip to content
Part IA Michaelmas, Lent, Easter Term

Series Convergence

Definitions

A series is k=0uk\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 limnSn=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 SSn<ε|S - S_n| < \varepsilon for all n>Nn > N.


Necessary Condition for Convergence

If uk\sum u_k converges, then limkuk=0\lim_{k \to \infty} u_k = 0.

Warning: The converse is false. Example: harmonic series 1k\sum \frac{1}{k} diverges even though 1k0\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 0ukvk0 \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 limkukvk=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=limkuk+1ukL = \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)212<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=11k=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. limkak=0\lim_{k \to \infty} a_k = 0

Example: The alternating harmonic series: k=1(1)k+1k=112+1314+=ln2\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=11kp\zeta(p) = \sum_{k=1}^{\infty} \frac{1}{k^p}

  • Converges for p>1p > 1
  • Diverges for p1p \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=limnanan+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=lim1/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.