91Ó°ÊÓ

The sine function, denoted as \( \sin(x) \), is one of the fundamental trigonometric functions, commonly appearing in various areas of mathematics, including calculus and sequences. It has a few noteworthy properties that are essential for solving problems like the one our textbook exercise presents.

The key properties to remember are:

• The sine function is periodic with a period of \(2\pi\), which means the sine curve repeats its pattern every \(2\pi\) radians.
• The range of \( \sin(x) \) is always within the closed interval \[ -1, 1 \]. Thus, for any real number \(x\), \( -1 \leq \sin(x) \leq 1 \).
• The sine graph is wave-like, oscillating between its peak of 1 and trough of -1.

Understanding these properties of the sine function helps solve problems involving trigonometric sequences. In the given exercise, recognizing that \( -1 \leq \sin(n^2) \leq 1 \) for any natural number \(n\) was the first pivotal step. Next, let's apply this knowledge to sequences and their behaviors in calculus.

Sequences in calculus are ordered lists of numbers following a specific rule, frequently expressed in terms of \(n\), where \(n\) is usually a positive integer called the index. They are tremendously important for understanding notions of convergence, limits, and the behavior of functions as they progress towards infinity.

In our textbook example, the sequence \(a_n = \frac{\sin(n^2)}{n+1}\) utilizes both the sine function and an indexing term. When dealing with sequences, one key question is their boundedness. A sequence is called bounded if there are real numbers \(M\) and \(m\) such that \(m \leq a_n \leq M\) for all \(n\) terms in the sequence.

To ascertain boundedness in this scenario, we can leverage our understanding of the sine function's limits. We know \(\sin(n^2)\) will always produce values within the interval \[ -1, 1 \], and as \(n\) gets larger, the denominator \(n+1\) also increases, effectively shrinking the overall value of \(a_n\) towards zero. This principle underpins the assertion that the sequence is indeed bounded, as no term in the series will ever breach the fixed range set by \(\sin(n^2)\) and the growing denominator.

Inequalities, particularly involving fractions, can be a bit tricky to navigate sometimes. However, they serve as a strong analytical tool for determining the behavior of sequences and functions, just as we did with our textbook exercise.

Let's delve a bit deeper into inequalities and fractions:

• When you have an inequality like \(\frac{a}{b} \leq \frac{c}{d}\), it indicates the fraction on the left side is less than or equal to the fraction on the right side, assuming \(b\) and \(d\) are positive.
• If you're dealing with negative denominators, remember that the inequality sign flips: \(\frac{a}{-b} \geq \frac{c}{-d}\).
• In the context of sequences, comparing \(\frac{\sin(n^2)}{n+1}\) with the fractions \(\frac{-1}{n+1}\) and \(\frac{1}{n+1}\), provides bounds for our sequence because of the sine function's range (\[ -1, 1 \]).

Applying this understanding to \(a_n\), we can confidently say the sequence is bounded because it will never exceed \(\frac{1}{n+1}\) nor fall below \(\frac{-1}{n+1}\). As \(n\) approaches infinity, both of these fractional bounds approach zero, affirming our conclusion that the sequence \(a_n\), behaves well within expected limits. This not only assures boundedness but can also guide predictions about the sequence's limit behavior.