Explanation Result used: Root test: Consider the series ∑ k = 1 ∞ a k . If lim k → ∞ | a k | k < 1 , then the series converges and if lim k → ∞ | a k | k > 1 , then the series diverges. Calculation: Simplify the series ∑ k = 1 ∞ ( k + a k ) k 2 , a > 0 as follows. ∑ k = 1 ∞ ( k + a k ) k 2 = ∑ k = 1 ∞ a k , where a k = ( k + a k ) k 2 . Compute the value of lim k → ∞ | a k | k . lim k → ∞ | a k | k = lim k → ∞ | ( k + a k ) k 2 | k =

Calculus, Single Variable: Early Transcendentals (3rd Edition)

3rd Edition

ISBN: 9780134766850

Author: William L. Briggs, Lyle Cochran, Bernard Gillett, Eric Schulz

Publisher: PEARSON

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Chapter 10.8, Problem 63E

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Whether the series ∑k=1∞(k+ak)k2,a>0 converges or diverges using the convergence test.

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Chapter 10.8, Problem 62E

Chapter 10.8, Problem 64E

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Chapter 10 Solutions

Calculus, Single Variable: Early Transcendentals (3rd Edition)

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Whether the series ∑ k = 1 ∞ ( k + a k ) k 2 , a > 0 converges or diverges using the convergence test.

Solution Summary: The author analyzes whether the series displaystyle 'underset' converges or diverges using the convergence test.

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Whether the series ∑k=1∞(k+ak)k2,a>0 converges or diverges using the convergence test.

Want to see the full answer?

Chapter 10 Solutions

Whether the series ∑ k = 1 ∞ ( k + a k ) k 2 , a &gt; 0 converges or diverges using the convergence test.

Solution Summary: The author analyzes whether the series displaystyle 'underset' converges or diverges using the convergence test.

Videos

Whether the series ∑k=1∞(k+ak)k2,a>0 converges or diverges using the convergence test.

Want to see the full answer?

Chapter 10 Solutions

Whether the series ∑ k = 1 ∞ ( k + a k ) k 2 , a > 0 converges or diverges using the convergence test.