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 convergent if 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 divergent then ∑ n = 1 ∞ b n is also divergent 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 an+1 ≤ an ∀ 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...

3rd Edition

ISBN: 9781464125263

Author: Jon Rogawski, Colin Adams

Publisher: W.H. Freeman & Co

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Chapter 11.4, Problem 38E

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To Prove:

If ${b_{n}}$ is a positive decreasing sequence with $\lim_{n \to \infty} b_{n} = 0$ , then the series $S = b_{1} + b_{2} - 2 b_{3} + a_{4} + b_{5} - 2 a_{6} + ...$ converges.

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Chapter 11.4, Problem 37E

Chapter 11.4, Problem 39E

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Chapter 11 Solutions

Calculus - Standalone book

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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To Prove: If { b n } is a positive decreasing sequence with lim n → ∞ b n = 0 , then the series S = b 1 + b 2 − 2 b 3 + a 4 + b 5 − 2 a 6 + ... converges.

Solution Summary: The author explains that if b n is a positive decreasing sequence, then the series converges.

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To Prove:

If ${b_{n}}$ is a positive decreasing sequence with $\lim_{n \to \infty} b_{n} = 0$ , then the series $S = b_{1} + b_{2} - 2 b_{3} + a_{4} + b_{5} - 2 a_{6} + ...$ converges.

Want to see the full answer?

Chapter 11 Solutions

To Prove: If { b n } is a positive decreasing sequence with lim n → ∞ b n = 0 , then the series S = b 1 + b 2 − 2 b 3 + a 4 + b 5 − 2 a 6 + ... converges.

Solution Summary: The author explains that if b n is a positive decreasing sequence, then the series converges.

Videos

To Prove: If {bn} is a positive decreasing sequence with limn→∞bn=0, then the series S=b1+b2−2b3+a4+b5−2a6+... converges.

Want to see the full answer?

Chapter 11 Solutions

To Prove:

If ${b_{n}}$ is a positive decreasing sequence with $\lim_{n \to \infty} b_{n} = 0$ , then the series $S = b_{1} + b_{2} - 2 b_{3} + a_{4} + b_{5} - 2 a_{6} + ...$ converges.