Definitions
A series is ∑ k = 0 ∞ u k \sum_{k=0}^{\infty} u_k ∑ k = 0 ∞ u k . The partial sums are S n = ∑ k = 0 n u k S_n = \sum_{k=0}^{n} u_k S n = ∑ k = 0 n u k .
Convergence: The series converges if lim n → ∞ S n = S \lim_{n \to \infty} S_n = S lim n → ∞ S n = S exists (finite). We write ∑ u k = S \sum u_k = S ∑ u k = S .
Formal definition: For any ε > 0 \varepsilon > 0 ε > 0 , there exists N N N such that ∣ S − S n ∣ < ε |S - S_n| < \varepsilon ∣ S − S n ∣ < ε for all n > N n > N n > N .
Necessary Condition for Convergence
If ∑ u k \sum u_k ∑ u k converges, then lim k → ∞ u k = 0 \lim_{k \to \infty} u_k = 0 lim k → ∞ u k = 0 .
Warning: The converse is false . Example: harmonic series ∑ 1 k \sum \frac{1}{k} ∑ k 1 diverges even though 1 k → 0 \frac{1}{k} \to 0 k 1 → 0 .
Absolute vs Conditional Convergence
Absolute convergence: ∑ ∣ u k ∣ \sum |u_k| ∑ ∣ u k ∣ converges.
Conditional convergence: ∑ u k \sum u_k ∑ u k converges but ∑ ∣ u k ∣ \sum |u_k| ∑ ∣ u k ∣ diverges.
Key theorem: If ∑ ∣ u k ∣ \sum |u_k| ∑ ∣ u k ∣ converges, then ∑ u k \sum u_k ∑ 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 ≤ u k ≤ v k 0 \leq u_k \leq v_k 0 ≤ u k ≤ v k :
If ∑ v k \sum v_k ∑ v k converges, then ∑ u k \sum u_k ∑ u k converges
If ∑ u k \sum u_k ∑ u k diverges, then ∑ v k \sum v_k ∑ v k diverges
Limit comparison: If lim k → ∞ u k v k = L \lim_{k \to \infty} \frac{u_k}{v_k} = L lim k → ∞ v k u k = L with 0 < L < ∞ 0 < L < \infty 0 < L < ∞ , then ∑ u k \sum u_k ∑ u k and ∑ v k \sum v_k ∑ v k behave the same.
Example: ∑ 1 k ( k + 1 ) \sum \frac{1}{k(k+1)} ∑ k ( k + 1 ) 1 converges since 1 k ( k + 1 ) < 1 k 2 \frac{1}{k(k+1)} < \frac{1}{k^2} k ( k + 1 ) 1 < k 2 1 and ∑ 1 k 2 \sum \frac{1}{k^2} ∑ k 2 1 converges.
Ratio Test
Let L = lim k → ∞ ∣ u k + 1 u k ∣ L = \lim_{k \to \infty} \left|\frac{u_{k+1}}{u_k}\right| L = lim k → ∞ u k u k + 1 :
L < 1 L < 1 L < 1 : series converges absolutely
L > 1 L > 1 L > 1 : series diverges
L = 1 L = 1 L = 1 : test is inconclusive
Example: ∑ k 2 2 k \sum \frac{k^2}{2^k} ∑ 2 k k 2 . Ratio: ( k + 1 ) 2 / 2 k + 1 k 2 / 2 k = 1 2 ( 1 + 1 k ) 2 → 1 2 < 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 k 2 / 2 k ( k + 1 ) 2 / 2 k + 1 = 2 1 ( 1 + k 1 ) 2 → 2 1 < 1 . Converges.
Harmonic Series Divergence
∑ k = 1 ∞ 1 k = 1 + 1 2 + 1 3 + 1 4 + ⋯ \sum_{k=1}^{\infty} \frac{1}{k} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots ∑ k = 1 ∞ k 1 = 1 + 2 1 + 3 1 + 4 1 + ⋯
Proof of divergence (grouping):
1 + 1 2 + ( 1 3 + 1 4 ) + ( 1 5 + 1 6 + 1 7 + 1 8 ) + ⋯ 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 + 2 1 + ( 3 1 + 4 1 ) + ( 5 1 + 6 1 + 7 1 + 8 1 ) + ⋯
> 1 + 1 2 + 1 2 + 1 2 + ⋯ > 1 + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \cdots > 1 + 2 1 + 2 1 + 2 1 + ⋯
The partial sums grow without bound.
Alternating Series (Leibniz Criterion)
An alternating series ∑ ( − 1 ) k a k \sum (-1)^k a_k ∑ ( − 1 ) k a k with a k > 0 a_k > 0 a k > 0 converges if:
a k a_k a k is monotonically decreasing for large k k k
lim k → ∞ a k = 0 \lim_{k \to \infty} a_k = 0 lim k → ∞ a k = 0
Example: The alternating harmonic series:
∑ k = 1 ∞ ( − 1 ) k + 1 k = 1 − 1 2 + 1 3 − 1 4 + ⋯ = ln 2 \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k} = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots = \ln 2 ∑ k = 1 ∞ k ( − 1 ) k + 1 = 1 − 2 1 + 3 1 − 4 1 + ⋯ = ln 2
This is conditionally convergent (not absolutely, since harmonic series diverges).
Riemann Zeta Function
ζ ( p ) = ∑ k = 1 ∞ 1 k p \zeta(p) = \sum_{k=1}^{\infty} \frac{1}{k^p} ζ ( p ) = ∑ k = 1 ∞ k p 1
Converges for p > 1 p > 1 p > 1
Diverges for p ≤ 1 p \leq 1 p ≤ 1
Special values:
ζ ( 2 ) = π 2 6 \zeta(2) = \frac{\pi^2}{6} ζ ( 2 ) = 6 π 2
ζ ( 4 ) = π 4 90 \zeta(4) = \frac{\pi^4}{90} ζ ( 4 ) = 90 π 4
Radius of Convergence for Power Series
For ∑ a n x n \sum a_n x^n ∑ a n x n , the radius of convergence is:
R = lim n → ∞ ∣ a n a n + 1 ∣ (if limit exists) R = \lim_{n \to \infty} \left|\frac{a_n}{a_{n+1}}\right| \quad \text{(if limit exists)} R = lim n → ∞ a n + 1 a n (if limit exists)
Converges absolutely for ∣ x ∣ < R |x| < R ∣ x ∣ < R
Diverges for ∣ x ∣ > R |x| > R ∣ x ∣ > R
At x = ± R x = \pm R x = ± R : must check separately
Example: ∑ x n n \sum \frac{x^n}{n} ∑ n x n .
R = lim 1 / n 1 / ( n + 1 ) = 1 R = \lim \frac{1/n}{1/(n+1)} = 1 R = lim 1/ ( n + 1 ) 1/ n = 1 . Converges for ∣ x ∣ < 1 |x| < 1 ∣ x ∣ < 1 .
At x = 1 x = 1 x = 1 : harmonic series diverges.
At x = − 1 x = -1 x = − 1 : alternating harmonic converges.
Summary: Choosing a Test
Situation Preferred Test Comparison with known series Comparison test Factorials, exponentials Ratio test Alternating signs Leibniz criterion Powers 1 / k p 1/k^p 1/ k p p-test (zeta) Power series Radius of convergence
Complexity: Ratio test is often easiest to apply. Comparison test requires knowing a suitable reference series.