Explanation If { a n } is a sequence of real numbers. Then the expression S n = a 1 + a 2 + ... + a n = ∑ k = 1 n a k is termed as sequence of n th partial sum, i.e., the sum of first n terms of a sequence and S = lim n → ∞ S n = a 1 + a 2 + ... + a n + ... = ∑ n = 1 ∞ a n is known as sum of infinite terms of sequence termed as infinite series. The given series is termed as positive term series if in the series S = a 1 + a 2 + ... + a n + ... = ∑ n = 1 ∞ a n , a n > 0 ∀ n ∈ N . An infinite series S = lim n → ∞ S n = a 1 + a 2 + ... + a n + ... = ∑ n = 1 ∞ a n is said to be convergent or divergent if its n th partial sum S n = a 1 + a 2 + ... + a n = ∑ k = 1 n a k is convergent to finite value or divergent to infinity. Necessary and sufficient condition for convergence Necessary and sufficient condition that infinite series S = ∑ n = 1 ∞ a n is said to be convergent if lim n → ∞ a n = 0 . Suppose a n ≥ 0 ∀ n ∈ N , then S = ∑ n = 1 ∞ a n is said to be convergentif and only if S n = a 1 + a 2 + ... + a n = ∑ k = 1 n a k is bounded above. Tests for Convergence Let us determine the convergence or the divergence of a series by comparing it to one whose behavior is already known. Comparison test: Suppose 0 ≤ a n ≤ b n ∀ n ≥ k for some k. Then (1) If ∑ n = 1 ∞ b n is convergent then ∑ n = 1 ∞ a n is also convergent (2) If ∑ n = 1 ∞ a n is divergentthen ∑ n = 1 ∞ b n is alsodivergent Limit Comparison Test: Suppose a n , b n ≥ 0 ∀ n ∈ N and lim n → ∞ a n b n = L (1) If L ∈ R , L > 0 , then both ∑ n = 1 ∞ a n and ∑ n = 1 ∞ b n converge or diverge together. (2) If L ∈ R , L = 0 and ∑ n = 1 ∞ b n converges, then ∑ n = 1 ∞ a n also converges (3) If L → ∞ and ∑ n = 1 ∞ b n diverges, then ∑ n = 1 ∞ a n also diverges Cauchy Test or Cauchy condensation Test : Suppose a n ≥ 0 ∀ n ∈ N and a n + 1 ≤ a n , then ∑ n = 1 ∞ a n converges if and only if ∑ k = 0 ∞ 2 k a 2 k converges. p- Series Test : (1) ∑ n = 1 ∞ 1 n p converges if p > 1 and diverges if p ≤ 1 (2) ∑ n = 2 ∞ 1 n ( log n ) p converges if p > 1 and diverges if p ≤ 1 Ratio Test: D’ Alembert Test: Suppose a n > 0 and lim n → ∞ a n a n + 1 = L . Then (1) If L ∈ R , L > 1 , then ∑ n = 1 ∞ a n converges. (2) If L ∈ R , L < 1 , then ∑ n = 1 ∞ a n diverges. (3) If L ∈ R , L = 1 , then test gets fail Raabe’s Test: Suppose a n > 0 and lim n → ∞ n ( a n a n + 1 − 1 ) = L . Then (1) If L ∈ R , L > 1 , then ∑ n = 1 ∞ a n converges. (2) If L ∈ R , L < 1 , then ∑ n = 1 ∞ a n diverges...

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ISBN: 9781464125263

Author: Jon Rogawski, Colin Adams

Publisher: W.H. Freeman & Co

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Chapter 11.4, Problem 34E

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Why the alternating series test is not applicable for given infinite series? The convergence of series by using alternate method.

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Chapter 11.4, Problem 33E

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Chapter 11 Solutions

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Ch. 11.1 - Prob. 1PQ Ch. 11.1 - Prob. 2PQ Ch. 11.1 - Prob. 3PQ Ch. 11.1 - Prob. 4PQ Ch. 11.1 - Prob. 5PQ Ch. 11.1 - Prob. 1E Ch. 11.1 - Prob. 2E Ch. 11.1 - Prob. 3E Ch. 11.1 - Prob. 4E Ch. 11.1 - Prob. 5E

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Why the alternating series test is not applicable for given infinite series? The convergence of series by using alternate method.

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Why the alternating series test is not applicable for given infinite series? The convergence of series by using alternate method.

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