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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 10.8.19

11–86. Applying convergence tests Determine whether the following series converge. Justify your answers.
∑ (from k = 0 to ∞) (3ᵏ⁺⁴) / (5ᵏ⁻²)

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Rewrite the general term of the series to simplify the expression. The term is given by \(\frac{3^{k+4}}{5^{k-2}}\). Use the properties of exponents to separate the powers of constants and the variable \(k\): \(\frac{3^{k} \cdot 3^{4}}{5^{k} \cdot 5^{-2}} = 3^{4} \cdot 5^{2} \cdot \frac{3^{k}}{5^{k}}\).
Simplify the term further by combining the powers with the same base in the numerator and denominator: \(3^{4} \cdot 5^{2} \cdot \left(\frac{3}{5}\right)^{k}\). This shows the term is a constant multiplied by a geometric term \(\left(\frac{3}{5}\right)^{k}\).
Recognize that the series is a geometric series of the form \(\sum_{k=0}^{\infty} ar^{k}\) where \(a = 3^{4} \cdot 5^{2}\) and \(r = \frac{3}{5}\). Recall that a geometric series converges if and only if \(|r| < 1\).
Check the value of the common ratio \(r = \frac{3}{5}\). Since \(\frac{3}{5} = 0.6\) and \(0.6 < 1\), the geometric series converges.
Conclude that the original series converges by the geometric series test because the common ratio's absolute value is less than 1.

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