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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 10.3.87e

87. Explain why or why not
Determine whether the following statements are true and give an explanation or counterexample.


e. ∑ (k = 1 to ∞) (π / e)⁻ᵏ is a convergent geometric series.

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Identify the general term of the series: the series is given by \( \sum_{k=1}^{\infty} \left( \frac{\pi}{e} \right)^{-k} \). This can be rewritten as \( \sum_{k=1}^{\infty} \left( \frac{e}{\pi} \right)^k \) because raising to the power \(-k\) is the same as taking the reciprocal and raising to the positive power \(k\).
Recognize that this is a geometric series with the first term \( a = \left( \frac{e}{\pi} \right)^1 = \frac{e}{\pi} \) and common ratio \( r = \frac{e}{\pi} \).
Recall the convergence criterion for a geometric series: a geometric series \( \sum a r^{k-1} \) converges if and only if \( |r| < 1 \).
Evaluate the absolute value of the common ratio: since \( e \approx 2.718 \) and \( \pi \approx 3.1415 \), \( \left| \frac{e}{\pi} \right| < 1 \).
Conclude that because \( |r| < 1 \), the series \( \sum_{k=1}^{\infty} \left( \frac{e}{\pi} \right)^k \) is a convergent geometric series.

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Geometric Series

A geometric series is a sum of terms where each term is found by multiplying the previous term by a constant ratio r. It has the form ∑ ar^k, where a is the first term and r is the common ratio. Understanding the structure helps identify if a series fits this pattern.
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Geometric Series

Convergence of Geometric Series

A geometric series converges if and only if the absolute value of the common ratio |r| is less than 1. When this condition holds, the infinite sum approaches a finite limit given by a/(1-r). If |r| ≥ 1, the series diverges.
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Geometric Series

Evaluating the Common Ratio

To determine convergence, it is essential to correctly identify and evaluate the common ratio r in the series. In this problem, the term (π/e)^(-k) can be rewritten as (e/π)^k, so the ratio is e/π. Comparing |e/π| to 1 determines if the series converges.
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Graphs of Common Functions