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Lecture 23Section 11.3 the Root Test; the Ratio Test

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Lecture 23Section 11.3 the Root Test; the Ratio Test 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 1 k −1 the series diverges since ak = 1 − k → e 6= 0. 3 3 The Ratio Test 3.1 The Ratio Test The Ratio Test: Comparison with Geometric Series Ratio Test ak+1 Suppose that ak > 0 for large k, and that [0.5ex] lim = λ for some λ > 0. k→∞ ak P • If λ < 1, then ak converges. P • If λ > 1, then ak diverges. • If λ = 1, then the test is inconclusive. Comparison with Geometric Series a • If P a is a multiple of a geometric series, e.g., P c λk, λ > 0, then k+1 k a P Pk is constant, i.e., λ. If λ < 1, then ak converges. If λ ≥ 1, then ak diverges. ak+1 k • If lim = λ < 1, then for large k, ak < cµ with λ < µ < 1. By the k→∞ a k P basic comparison test, ak converges. Examples X k2 converges, by the ratio test: 2k 2 2 k 3 ak+1 (k + 1) k 2 (k + 1) = k+1 ÷ k = k+1 · 3 ak 2 2 2 k 1 (k + 1)3 1 = · → as k → ∞ 2 k3 2 X 1 converges, by the ratio test: k! a 1 1 k! k+1 = ÷ = ak (k + 1)! k! (k + 1)! 1 = → 0 as k → ∞ k + 1 4 Examples X k converges, by the ratio test: 10k k ak+1 k + 1 k 10 k + 1 = k+1 ÷ k = k+1 · ak 10 10 10 k 1 k + 1 1 = · → as k → ∞ 10 k 10 X kk diverges, by the ratio test: k! k+1 k k+1 k ak+1 (k + 1) k k! (k + 1) (k + 1) = ÷ = · k = k ak (k + 1)! k! (k + 1)! k k k + 1k 1 k = = 1 + → e > 1 as k → ∞ k k Examples X 2k converges, by the ratio test: 3k − 2k k+1 k k+1 k k ak+1 2 2 2 3 − 2 = k+1 k+1 ÷ k k = k · k+1 k+1 ak 3 − 2 3 − 2 2 3 − 2 1 − (2/3)k 1 = 2 · → 2 · < 1 as k → ∞ 3 − 2(2/3)k 3 X 1 √ converges, by the ratio test: k! s ak+1 1 1 k! = p ÷ √ = ak (k + 1)! k! (k + 1)! r 1 = → 0 as k → ∞ k + 1 Examples X k22k converges, by the ratio test: k! 2 k+1 2 k 2 k+1 ak+1 (k + 1) 2 k 2 (k + 1) 2 k! = ÷ = 2 · k · ak (k + 1)! k! k 2 (k + 1)! (k + 1)2 2 = · → 0 < 1 as k → ∞ k2 k + 1 5 X 1 inconclusive, by the ratio test: 2k + 1 a 1 1 2k + 1 2k + 1 k+1 = ÷ = = ak 2(k + 1) + 1 2k + 1 2(k + 1) + 1 2k + 3 2 + 1/k = → 1 as k → ∞ 2 + 3/k [2ex] the series diverges by the limit comparison test: 1 1 2k X 1 ÷ = → 1 as k → ∞ and diverges. 2k + 1 2k 2k + 1 2k Outline Contents 1 Comparison Tests 1 2 Root Test 2 2.1 Root Test . 2 3 Ratio Test 4 3.1 Ratio Test . 4 6.
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