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 the 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 . The infinite geometric series S = ∑ n = 1 ∞ a r n , where a is the first term and r is the common ratio is (i) Convergent if | r | < 1 (ii) Divergent if | r | ≥ 1 (iii) Alternating series if r = − 1 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...

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Author: Rogawski

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Chapter 11, Problem 28CRE

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The sum of given infinite series.

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Chapter 11, Problem 27CRE

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Chapter 11 Solutions

CALCULUS W/SAPLING ACCESS >IC<

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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The sum of given infinite series.

Solution Summary: The author explains that a sequence of real numbers is termed as the sum of n th partial sum. An infinite series is called convergent or divergent.

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The sum of given infinite series.

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Chapter 11 Solutions