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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:
  • 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.
In our specific case with the sequence \( a_n = \frac{\sin n}{n} \), we know \(-1 \leq \sin n \leq 1\). Dividing the whole inequality by \( n \), it becomes \(-\frac{1}{n} \leq \frac{\sin n}{n} \leq \frac{1}{n}\). Both \(-\frac{1}{n}\) and \(\frac{1}{n}\) go to zero as \( n \) approaches infinity, so by the Squeeze Theorem, \( \frac{\sin n}{n} \) also converges to zero.
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:
  • The numerator, \( \sin n \), remains bounded between -1 and 1.
  • The denominator, \( n \), grows indefinitely.
Dividing by an ever-increasing number tends to "shrink" the fraction towards zero. Hence, the sequence \( \frac{\sin n}{n} \) converges to the limit 0.
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.

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