Chapter 9: Problem 23
Do the sequences, converge or diverge? If a sequence converges, find its limit. $$\frac{\sin n}{n}$$
Short Answer
Expert verified
The sequence converges to 0.
Step by step solution
01
Understanding the Sequence
We are given the sequence \( a_n = \frac{\sin n}{n} \). We need to determine whether this sequence converges or diverges and find its limit if it converges.
02
Analyze the Behavior of \( \sin n \)
Recall that the function \( \sin n \) oscillates between -1 and 1 for any integer \( n \). This means that the values of \( \sin n \) remain bounded within this range.
03
Consider the Denominator \( n \)
Notice that as \( n \to \infty \), the denominator \( n \) increases without bound. Any function or number divided by an increasingly larger number tends to zero.
04
Applying the Squeeze Theorem
We know \( -1 \leq \sin n \leq 1 \). Dividing the inequality throughout by \( n \), we get \( -\frac{1}{n} \leq \frac{\sin n}{n} \leq \frac{1}{n} \). As \( n \to \infty \), both \( -\frac{1}{n} \) and \( \frac{1}{n} \) approach zero.
05
Conclude with Squeeze Theorem
Since \( \lim_{n \to \infty} -\frac{1}{n} = 0 \) and \( \lim_{n \to \infty} \frac{1}{n} = 0 \), the Squeeze Theorem implies that \( \lim_{n \to \infty} \frac{\sin n}{n} = 0 \). Thus, the sequence \( \frac{\sin n}{n} \) converges to 0.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Squeeze Theorem
The Squeeze Theorem is a fundamental concept used in calculus, especially useful when dealing with sequences and limits. In simple terms, the Squeeze Theorem states that if you have two sequences or functions "squeezing" a third sequence/function between them, and they all converge to the same limit, then the middle one must converge to that limit as well. This technique is particularly helpful when a sequence itself is complicated to directly deal with.
To apply the Squeeze Theorem:
To apply the Squeeze Theorem:
- You first establish two simpler sequences or functions that "sandwich" the given sequence.
- Show that both these sequences converge to the same limit as the bigger sequence.
- Finally, conclude that your given sequence converges to that same limit.
Limit of a Sequence
The concept of the limit of a sequence plays a crucial role in determining whether a sequence converges or diverges. A sequence converges if its terms approach a specific number, called the limit, as you move to infinity. Conversely, if a sequence does not approach any particular number, it diverges.
For sequences, the notation \( \lim_{n \to \infty} a_n = L\) indicates that as \( n \) becomes very large, the terms \( a_n \) come close to \( L\), the limit.
In the sequence \( \frac{\sin n}{n} \), we see that as \( n \) increases:
For sequences, the notation \( \lim_{n \to \infty} a_n = L\) indicates that as \( n \) becomes very large, the terms \( a_n \) come close to \( L\), the limit.
In the sequence \( \frac{\sin n}{n} \), we see that as \( n \) increases:
- The numerator, \( \sin n \), remains bounded between -1 and 1.
- The denominator, \( n \), grows indefinitely.
Oscillating Functions
Oscillating functions are functions that move between two values repeatedly and do not settle at a particular point as their input changes. The sine function, \( \sin n \), is a classic example of an oscillating function. It varies between -1 and 1 for all integer inputs, creating waves of regular intervals.
When an oscillating function, like \( \sin n \), is part of a sequence such as \( \frac{\sin n}{n} \), its endless ups and downs can complicate convergence analysis at first glance. However, because we know the oscillation is limited to a certain range, it makes it easier to use tools like the Squeeze Theorem to analyze its behavior as part of more complex expressions.
In convergence discussions, understanding that \( \sin n \) itself remains limited helps us reason that multiplying or dividing such behavior by other terms (like \( n \)) can result in a more manageable function or sequence. Hence, even though \( \sin n \) oscillates, \( \frac{\sin n}{n} \) does not oscillate towards infinity. Instead, it converges to a stable limit, which is zero in this case.
When an oscillating function, like \( \sin n \), is part of a sequence such as \( \frac{\sin n}{n} \), its endless ups and downs can complicate convergence analysis at first glance. However, because we know the oscillation is limited to a certain range, it makes it easier to use tools like the Squeeze Theorem to analyze its behavior as part of more complex expressions.
In convergence discussions, understanding that \( \sin n \) itself remains limited helps us reason that multiplying or dividing such behavior by other terms (like \( n \)) can result in a more manageable function or sequence. Hence, even though \( \sin n \) oscillates, \( \frac{\sin n}{n} \) does not oscillate towards infinity. Instead, it converges to a stable limit, which is zero in this case.