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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 10.2.83f

Explain why or why not
Determine whether the following statements are true and give an explanation or counterexample.
f.If the sequence {aₙ} diverges, then the sequence {0.000001aₙ} diverges.

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1
Recall the definition of divergence for a sequence: a sequence \( \{a_n\} \) diverges if it does not approach a finite limit as \( n \to \infty \).
Consider the sequence \( \{0.000001 \cdot a_n\} \). This is the original sequence \( \{a_n\} \) multiplied by a constant scalar \( 0.000001 \).
Multiplying a sequence by a nonzero constant scales its terms but does not change whether the sequence converges or diverges. Specifically, if \( \{a_n\} \) diverges to infinity or oscillates without limit, then \( \{0.000001 \cdot a_n\} \) will also diverge (though possibly to a different infinite value or oscillation).
However, if \( \{a_n\} \) diverges because it oscillates or does not settle to a limit, scaling by \( 0.000001 \) will not make it converge; it will still fail to approach a finite limit.
Therefore, the statement is true: if \( \{a_n\} \) diverges, then \( \{0.000001 \cdot a_n\} \) also diverges.

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