55–70. More sequences Find the limit of the following sequences or determine that the sequence diverges.
{sinn / 2ⁿ}
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Identify the given sequence as \( a_n = \frac{\sin n}{2^n} \), where \( n \) is a positive integer.
Recall that \( \sin n \) oscillates between -1 and 1 for all integer values of \( n \), so \( \sin n \) is bounded.
Note that the denominator \( 2^n \) is an exponential function that grows without bound as \( n \to \infty \).
Since the numerator is bounded and the denominator grows exponentially, the fraction \( \frac{\sin n}{2^n} \) approaches zero as \( n \to \infty \).
Conclude that the limit of the sequence \( \left\{ \frac{\sin n}{2^n} \right\} \) is zero.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Sequences and Limits
A sequence is an ordered list of numbers defined by a specific rule. The limit of a sequence is the value that the terms approach as the index goes to infinity. Understanding how to determine if a sequence converges (has a limit) or diverges (does not have a limit) is fundamental in calculus.
Exponential functions like 2ⁿ grow very rapidly as n increases. When a sequence has terms divided by an exponential function with base greater than 1, the denominator grows faster than the numerator, often causing the sequence to approach zero. This property helps in evaluating limits involving exponential terms.
The sine function oscillates between -1 and 1 for all real numbers. This boundedness means that sin(n) remains within fixed limits regardless of n. When combined with a rapidly growing denominator, the bounded numerator ensures the sequence terms become very small, aiding in limit evaluation.