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Ch. 10 - Sequences and Infinite Series
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Non è quello che usi tu?Cambia libro di testo
Capitolo 10, Problema 10.3.85

72–86. Evaluating series Evaluate each series or state that it diverges.
∑ (k = 1 to ∞) (((1/6)ᵏ + (1/3)ᵏ) × k⁻¹)

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Identify the given series: \( \sum_{k=1}^{\infty} \left( \left( \frac{1}{6} \right)^k + \left( \frac{1}{3} \right)^k \right) \times \frac{1}{k} \). This is a sum of two series combined inside the summation.
Rewrite the series by separating the sum into two separate series: \( \sum_{k=1}^{\infty} \frac{\left( \frac{1}{6} \right)^k}{k} + \sum_{k=1}^{\infty} \frac{\left( \frac{1}{3} \right)^k}{k} \). This allows us to analyze each series individually.
Recognize that each series is of the form \( \sum_{k=1}^{\infty} \frac{x^k}{k} \), which is related to the Taylor series expansion of the function \( -\ln(1 - x) \) for \( |x| < 1 \).
Check the convergence criteria: since \( \left| \frac{1}{6} \right| < 1 \) and \( \left| \frac{1}{3} \right| < 1 \), both series converge absolutely.
Express each series using the logarithmic function: \( \sum_{k=1}^{\infty} \frac{x^k}{k} = -\ln(1 - x) \). Substitute \( x = \frac{1}{6} \) and \( x = \frac{1}{3} \) respectively, then add the two results to write the sum in closed form.

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