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Ch. 10 - Infinite Sequences and Series
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Non è quello che usi tu?Cambia libro di testo
Capitolo 10, Problema 10.5.66a

Assume that bₙ is a sequence of positive numbers converging to 1/3. Determine if the following series converge or diverge.
a. ∑ (from n = 1 to ∞) [(bₙ₊₁ + bₙ) / n 4ⁿ]

Guida verificata passo dopo passo
1
Identify the general term of the series: \( a_n = \frac{b_{n+1} + b_n}{n 4^n} \). Since \( b_n \) converges to \( \frac{1}{3} \), both \( b_n \) and \( b_{n+1} \) approach \( \frac{1}{3} \) as \( n \to \infty \).
Use the limit of \( b_n \) to approximate the behavior of the numerator for large \( n \): \( b_{n+1} + b_n \to \frac{1}{3} + \frac{1}{3} = \frac{2}{3} \). This helps simplify the term for large \( n \).
Focus on the denominator \( n 4^n \), which grows very rapidly due to the exponential term \( 4^n \). This suggests the terms \( a_n \) decrease quickly.
Apply the Comparison Test or Limit Comparison Test by comparing \( a_n \) to a known convergent series, such as \( \frac{1}{4^n} \), since \( \frac{b_{n+1} + b_n}{n 4^n} \) behaves similarly to \( \frac{constant}{n 4^n} \).
Conclude about convergence: since \( \sum \frac{1}{4^n} \) converges (geometric series with ratio \( \frac{1}{4} < 1 \)) and \( \frac{1}{n} \) grows slowly, the given series converges by comparison.

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