Problem 46 - Thomas Calculus

Question

Which of the sequences \(\left\\{a_{n}\right\\}\) in Exercises \(27-90\) converge, and which diverge? Find the limit of each convergent sequence. $$ a_{n}=\frac{\sin ^{2} n}{2^{n}} $$

Step-by-Step Solution

Answer

The sequence converges to 0.

Step 1: Define the Sequence

The sequence given is \(a_n = \frac{\sin^2 n}{2^n}\), where \(n\) is a positive integer.

Step 2: Analyze the Numerator

The term \(\sin^2 n\) is bounded because the sine function oscillates between -1 and 1. This means \(\sin^2 n\) is always between 0 and 1, inclusive.

Step 3: Analyze the Denominator

The denominator is \(2^n\), an exponential function that increases very rapidly as \(n\) increases. Exponential functions grow faster than any polynomial or bounded function.

Step 4: Examine the Behavior as \(n\) Increases

As \(n\) becomes large, the term \( \frac{\sin^2 n}{2^n} \) is the quotient of a bounded numerator (bounded by 1) and a rapidly growing denominator.

Step 5: Apply the Squeeze Theorem

Since \(0 \leq \sin^2 n \leq 1\), we have \(0 \leq \frac{\sin^2 n}{2^n} \leq \frac{1}{2^n}\). The right side, \(\frac{1}{2^n}\), tends to 0 as \(n\) goes to infinity. By the Squeeze Theorem, \(a_n\) also tends to 0.

Step 6: Conclusion on Convergence

The sequence converges to 0, because the term \( \frac{\sin^2 n}{2^n} \) approaches 0 as \(n\) tends to infinity, based on our calculations and the Squeeze Theorem.

Key Concepts

Squeeze Theorem

The Squeeze Theorem is an incredibly useful tool in calculus when it comes to determining the limits of sequences. It essentially allows us to 'squeeze' a sequence between two other sequences whose limits we already understand. In the example of the sequence \(a_n = \frac{\sin^2 n}{2^n}\), we can apply this theorem to find its limit.Here's how it works:

• We establish an upper and a lower bound for our sequence. In this case, because \(0 \leq \sin^2 n \leq 1\), we know \(0 \leq \frac{\sin^2 n}{2^n} \leq \frac{1}{2^n}\).
• The upper bound, \(\frac{1}{2^n}\), is a sequence that converges to zero as \(n\) becomes very large.

These observations allow us to conclude that \(\frac{\sin^2 n}{2^n}\) will consequently converge to zero by the Squeeze Theorem, since it cannot exceed \(\frac{1}{2^n}\) which approaches zero.
This theorem is particularly powerful for solving limits where direct evaluation is challenging. By creatively finding bounds, we can deduce the behavior of complex sequences. It's a nifty trick that often simplifies seemingly daunting problems.

Exponential Growth

Understanding exponential growth is key to analyzing the sequence given. Exponential growth refers to the scenario where a quantity increases by a constant percentage over equal time intervals, and is represented in our sequence by \(2^n\). This type of growth leads to numbers that increase very quickly as the exponent, \(n\), rises.
Let's break down why this matters:

• For the sequence \(a_n = \frac{\sin^2 n}{2^n}\), despite the numerator being bounded by 1, the term \(2^n\) in the denominator increases exponentially.
• This rapid increase—growing faster than any simple oscillating function like sine—ensures that \(\frac{1}{2^n}\) approaches zero as \(n\) tends to infinity.

In terms of exponential growth, it's important to remember that exponential functions will "outpace" other components of a sequence, leading to a more significant effect on the limit. Recognizing this nature helps in understanding why certain sequences converge to zero when divided by an exponentially growing sequence.

Numerical Limits

Numerical limits are foundational in the convergence of sequences. They provide the scaffolding to understand where a sequence like \(a_n = \frac{\sin^2 n}{2^n}\) is heading as \(n\) increases towards infinity.
To deepen our understanding:

• A numerical limit of a sequence \(\{a_n\}\) is the value that \(a_n\) approaches as \(n\) grows without bound.
• In our example, the numerical limit is 0, set by the behavior of the denominator, \(2^n\), overcoming the oscillations of \(\sin^2 n\).

When studying limits, the bounds and behaviors of both the numerator and the denominator are crucial. Since \(\sin^2 n\) is bounded and \(2^n\) grows rapidly, the sequence "runs" towards zero, giving us a concrete numerical limit. Such analysis is the cornerstone of determining sequence convergence and finding limits effectively.