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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 10.1.48

Convergence and Divergence
Which of the sequences {aₙ} in Exercises 31–100 converge, and which diverge? Find the limit of each convergent sequence.
aₙ = nπ cos(nπ)

Guida verificata passo dopo passo
1
Start by analyzing the given sequence: \(a_n = n\pi \cos(n\pi)\). Notice that \(n\pi\) is a linear term in \(n\), and \(\cos(n\pi)\) is a trigonometric term that depends on \(n\).
Recall the behavior of \(\cos(n\pi)\). Since \(\cos(\theta)\) has period \(2\pi\), evaluate \(\cos(n\pi)\) for integer values of \(n\). Specifically, \(\cos(n\pi) = (-1)^n\) because \(\cos(n\pi)\) alternates between 1 and -1 depending on whether \(n\) is even or odd.
Rewrite the sequence using this identity: \(a_n = n\pi (-1)^n\). This means the sequence terms are \(n\pi\) multiplied by either 1 or -1, alternating sign as \(n\) increases.
Consider the limit of \(a_n\) as \(n\) approaches infinity. Since \(n\pi\) grows without bound and \((-1)^n\) only changes the sign, the terms oscillate between large positive and large negative values, so the sequence does not approach a finite limit.
Conclude that the sequence \(a_n\) diverges because it does not settle to a single finite value as \(n\) becomes very large.

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