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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 10.8.15

11–86. Applying convergence tests Determine whether the following series converge. Justify your answers.
∑ (from k = 1 to ∞) (−7)ᵏ / k!

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Identify the given series: \( \sum_{k=1}^{\infty} \frac{(-7)^k}{k!} \). This is an infinite series with terms involving factorials in the denominator.
Recall that factorials grow very rapidly, which often suggests the series might converge. To confirm, consider applying the Ratio Test, which is useful for series with factorials and exponentials.
Set up the Ratio Test by examining the limit \( L = \lim_{k \to \infty} \left| \frac{a_{k+1}}{a_k} \right| \), where \( a_k = \frac{(-7)^k}{k!} \). Substitute to get \( L = \lim_{k \to \infty} \left| \frac{(-7)^{k+1} / (k+1)!}{(-7)^k / k!} \right| \).
Simplify the expression inside the limit: \( L = \lim_{k \to \infty} \left| \frac{-7}{k+1} \right| = \lim_{k \to \infty} \frac{7}{k+1} \). Since \( \frac{7}{k+1} \to 0 \) as \( k \to \infty \), the limit \( L = 0 \).
Interpret the Ratio Test result: since \( L < 1 \), the series \( \sum_{k=1}^{\infty} \frac{(-7)^k}{k!} \) converges absolutely. Therefore, the series converges.

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