Growth rates of sequences Use Theorem 10.6 to find the limit of the following sequences or state that they diverge.
{n¹⁰⁰⁰ / 2ⁿ}
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Identify the sequence given: \(a_n = \frac{n^{1000}}{2^n}\).
Recall Theorem 10.6, which states that for sequences of the form \(\frac{n^p}{r^n}\) where \(p\) is a positive integer and \(r > 1\), the exponential term in the denominator grows faster than any polynomial term in the numerator.
Apply this theorem to the sequence: since \$2^n\( grows exponentially and \)n^{1000}$ grows polynomially, the denominator grows faster than the numerator as \(n \to \infty\).
Conclude that the limit of the sequence \(a_n\) as \(n\) approaches infinity is 0 because the denominator dominates the numerator.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Limits of Sequences
A limit of a sequence describes the value that the terms of the sequence approach as the index goes to infinity. Understanding how to evaluate these limits helps determine whether a sequence converges to a finite number or diverges to infinity or does not settle at all.
Comparing growth rates involves analyzing how fast different functions increase as their input becomes very large. For example, exponential functions like 2^n grow faster than any polynomial function like n^1000, which is crucial in determining the behavior of sequences involving these terms.
Theorem 10.6 typically refers to a result that helps find limits of sequences by comparing them to known sequences with established limits. It often involves using dominant terms or applying limit laws to simplify complex sequences and determine their convergence or divergence.