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Ch. 10 - Sequences and Infinite Series
Briggs - Calculus: Early Transcendentals 3rd Edition
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Tutti i libri di testoBriggs 3rd EditionCh. 10 - Sequences and Infinite SeriesProblema 10.5.5
Capitolo 10, Problema 10.5.5

What comparison series would you use with the Comparison Test to determine whether ∑ (k = 1 to ∞) 2ᵏ / (3ᵏ + 1) converges?

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Identify the general term of the series: \(a_k = \frac{2^k}{3^k + 1}\).
To apply the Comparison Test, find a simpler series \(b_k\) that resembles \(a_k\) for large \(k\) and whose convergence behavior is known.
Since \$3^k$ dominates the $+1$ in the denominator for large $k$, approximate $a_k$ by \(\frac{2^k}{3^k}\).
Simplify the approximation: \(\frac{2^k}{3^k} = \left(\frac{2}{3}\right)^k\).
Use the geometric series \(\sum_{k=1}^\infty \left(\frac{2}{3}\right)^k\) as the comparison series, because it converges (common ratio \(\frac{2}{3} < 1\)).

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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; if larger than those of a divergent series, it diverges.
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Direct Comparison Test

Geometric Series

A geometric series has terms of the form ar^k, where r is the common ratio. It converges if |r| < 1 and diverges otherwise. Recognizing geometric behavior helps in choosing a suitable comparison series for convergence tests.
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Geometric Series

Asymptotic Behavior of Terms

Analyzing the dominant terms in the numerator and denominator for large k helps simplify the series term. For example, 2^k/(3^k + 1) behaves like (2/3)^k for large k, guiding the choice of a comparison series with similar asymptotic behavior.
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Asymptotes of Hyperbolas
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