Multiple Choice
Determine the convergence or divergence of the series.
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14. Sequences & Series
Convergence Tests
1:34 minutesProblem 10.5.3
Textbook Question
What comparison series would you use with the Comparison Test to determine whether
∑ (k = 1 to ∞) 1 / (k² + 1) converges?
Verified step by step guidance1
Identify the given series: \( \sum_{k=1}^{\infty} \frac{1}{k^2 + 1} \). We want to determine if it converges using the Comparison Test.
Recall that for large values of \( k \), the term \( \frac{1}{k^2 + 1} \) behaves similarly to \( \frac{1}{k^2} \) because the \( +1 \) becomes insignificant compared to \( k^2 \).
Choose a comparison series that is simpler but related, such as \( \sum_{k=1}^{\infty} \frac{1}{k^2} \), which is a p-series with \( p = 2 > 1 \) and is known to converge.
Use the Comparison Test by noting that \( \frac{1}{k^2 + 1} < \frac{1}{k^2} \) for all \( k \geq 1 \), so if the larger series \( \sum \frac{1}{k^2} \) converges, then the original series also converges.
Conclude that the appropriate comparison series to use is \( \sum_{k=1}^{\infty} \frac{1}{k^2} \) because it helps establish convergence of the original series by the Comparison Test.
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Video duration:
1mKey Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Comparison Test
The Comparison Test determines the convergence or divergence of a series by comparing it to another series with known behavior. If the terms of the given series are smaller than those of a convergent series, it also converges. Conversely, if the terms are larger than those of a divergent series, it diverges.
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Choice of Comparison Series
To apply the Comparison Test effectively, select a simpler series whose convergence is known and whose terms closely resemble the original series. For ∑ 1/(k² + 1), the series ∑ 1/k² is a natural choice because for large k, 1/(k² + 1) behaves like 1/k².
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Convergence of p-Series
A p-series ∑ 1/k^p converges if and only if p > 1. Since ∑ 1/k² is a p-series with p = 2, it converges. This fact helps conclude that ∑ 1/(k² + 1) also converges by comparison.
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