55–70. More sequences Find the limit of the following sequences or determine that the sequence diverges.
aₙ = e⁻ⁿcosn
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Identify the given sequence: \(a_n = e^{-n} \cos n\).
Recall that \(e^{-n}\) represents an exponential decay term, which approaches 0 as \(n\) approaches infinity.
Note that \(\cos n\) oscillates between -1 and 1 for all integer values of \(n\) and does not have a limit.
Since \(e^{-n}\) tends to 0 and \(\cos n\) is bounded, the product \(e^{-n} \cos n\) will be squeezed between \(-e^{-n}\) and \(e^{-n}\).
Apply the Squeeze Theorem: because both \(-e^{-n}\) and \(e^{-n}\) approach 0, the sequence \(a_n = e^{-n} \cos n\) also approaches 0 as \(n\) approaches infinity.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Limits of Sequences
The limit of a sequence is the value that the terms of the sequence approach as the index n goes to infinity. If the terms get arbitrarily close to a fixed number, the sequence converges; otherwise, it diverges. Understanding limits helps determine the long-term behavior of sequences.
Exponential decay refers to functions of the form e^(-n), which decrease rapidly towards zero as n increases. In sequences, this factor often dominates the behavior, causing terms to shrink and potentially leading the sequence to converge to zero.
Functions like cos(n) oscillate between -1 and 1 without settling to a single value. When combined with other factors, such as exponential decay, the oscillations influence the sequence's behavior but may be dampened if multiplied by a term tending to zero.