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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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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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Sequence Divergence
A sequence diverges if it does not approach a finite limit as n approaches infinity. Divergence means the terms either grow without bound, oscillate, or fail to settle at any single value.
Multiplying each term of a sequence by a constant scales the sequence but does not necessarily preserve its convergence or divergence properties. The behavior depends on the original sequence and the scalar.
A counterexample disproves a universal statement by providing a specific case where the statement fails. Using counterexamples is essential to test the validity of claims about sequences.