Textbook Question
11–86. Applying convergence tests Determine whether the following series converge. Justify your answers.
∑ (from k = 1 to ∞)(cos(1 / k) – cos(1 / (k + 1)))
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14. Sequences & Series
Convergence Tests
2:29 minutesProblem 10.8.71
Textbook Question
11–86. Applying convergence tests Determine whether the following series converge. Justify your answers.
∑ (from k = 1 to ∞)(ln²k) / k³ᐟ²
Verified step by step guidance1
Identify the general term of the series: \(a_k = \frac{(\ln k)^2}{k^{3/2}}\).
Recall that for series with positive terms, common convergence tests include the Comparison Test, the Limit Comparison Test, and the p-series test.
Note that the denominator \(k^{3/2}\) is a power function with exponent \(\frac{3}{2} > 1\), which suggests the series resembles a p-series \(\sum \frac{1}{k^p}\) with \(p = \frac{3}{2}\).
Since \(\ln^2 k\) grows slower than any positive power of \(k\), compare \(a_k\) to \(\frac{1}{k^{3/2}}\) using the Limit Comparison Test: compute \(\lim_{k \to \infty} \frac{a_k}{1/k^{3/2}} = \lim_{k \to \infty} (\ln k)^2\).
Because \(\lim_{k \to \infty} (\ln k)^2 = \infty\), the Limit Comparison Test with \$1/k^{3/2}\( is inconclusive; instead, use the Comparison Test noting that for large \)k\(, \((\ln k)^2\) grows slower than any power \(k^{\epsilon}\) for \(\epsilon > 0\), so \)a_k$ behaves like \(\frac{1}{k^{3/2 - \epsilon}}\) which still converges since \(3/2 - \epsilon > 1\). Therefore, the series converges by comparison to a convergent p-series.
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Video duration:
2mKey Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Convergence of Infinite Series
An infinite series converges if the sequence of its partial sums approaches a finite limit. Determining convergence involves analyzing the behavior of the terms as the index grows large, ensuring the sum does not diverge to infinity.
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Convergence of an Infinite Series
Comparison and Limit Comparison Tests
These tests compare a given series to a known benchmark series to determine convergence. The Comparison Test uses inequalities, while the Limit Comparison Test uses the limit of the ratio of terms to decide if both series share the same convergence behavior.
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Behavior of Logarithmic and Power Functions in Series
Understanding how logarithmic terms like (ln k)² grow relative to polynomial terms like k^(3/2) is crucial. Since polynomial terms grow faster than logarithmic terms, this relationship helps in applying comparison tests to determine series convergence.
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