Lecture 23Section 11.3 The Root Test; The Ratio Test
Jiwen He
1 Comparison Tests
Basic Series that Converge or Diverge ∞ ∞ X X ak converges iff ak converges, ∀j ≥ 1. k=1 k=j In determining whether a series converges, it does not matter where the sum- P mation begins. Thus, we will omit it and write ak. Basic Series that Converge X Geometric series: xk, if |x| < 1 X 1 p-series: , if p > 1 kp
Basic Series that Diverge X Any series ak for which lim ak 6= 0 k→∞ X 1 p-series: , if p ≤ 1 kp
Quiz Quiz X 1 1. (a) converges, (b) diverges. n
X 1 2. √ (a) converges, (b) diverges. n
X 1 1. Harmonic series diverges. n
X 1 1 2. √ p-series with p = diverges. n 2
1 Comparison Tests Basic Comparison Test Suppose that 0 ≤ ak ≤ bk for sufficiently large k. X X If bk converges, then so does ak. X X If ak diverges, then so does bk.
Limit Comparison Test ak Suppose that ak > 0 and bk > 0 for sufficiently large k, and that lim = L k→∞ bk for some L > 0. X X ak converges iff bk converges.
Quiz Quiz
X 1 3. (a) converges, (b) diverges. n2 + 1
X 2n 4. √ (a) converges, (b) diverges. 3n3 + 5
X 1 X 1 3. converges by comparison with . n2 + 1 n2
X 2n X 2 4. √ converges by comparison with . 3n3 + 5 3n2
2 The Root Test 2.1 The Root Test The Root Test: Comparison with Geometric Series Root Test 1 Suppose that ak > 0 for large k, and that [0.5ex] lim (ak) k = ρ for some ρ > 0. k→∞ P • If ρ < 1, then ak converges. P • If ρ > 1, then ak diverges. • If ρ = 1, then the test is inconclusive.
2 Comparison with Geometric Series
P P k 1 • If ak is a geometric series, e.g., ρ , ρ > 0, then (ak) k is constant, P P i.e., ρ. If ρ < 1, then ak converges. If ρ ≥ 1, then ak diverges.
1 k • If lim (ak) k = ρ < 1, then for large k, ak < µ with ρ < µ < 1. By the k→∞ P basic comparison test, ak converges.
Examples X k2 converges, by the root test: 2k 2 1/k k 1 1/k (a )1/k = = · k2 k 2k 2 1 h i2 1 = · k1/k → · 1 < 1 as k → ∞ 2 2
X 1 converges, by the root test: (ln k)k 1 1/k (a )1/k = k (ln k)k 1 = → 0 as k → ∞ ln k
Examples
k2 X 1 1 − converges, by the root test: k 1 k (−1)k (a )1/k = 1 − = 1 + → e−1 < 1 as k → ∞ k k k
k X 1 1 − inconclusive, by the root test: k 1 (a )1/k = 1 − → 1 as k → ∞ k k