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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 10.2.67

55–70. More sequences
Find the limit of the following sequences or determine that the sequence diverges.


{sinn / 2ⁿ}

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1
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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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.
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Introduction to Sequences

Behavior of Exponential Functions

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.
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Graphs of Exponential Functions

Boundedness of the Sine Function

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.
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Graph of Sine and Cosine Function