Problem 43 - Calculus: Early Transcendentals

Question

Determine whether the sequence converges or diverges. If it converges, find the limit. \( a_n = \frac { \cos^2 n}{2^n} \)

Step-by-Step Solution

Answer

The sequence converges to 0.

Step 1: Understand the Sequence

The sequence given is \( a_n = \frac {\cos^2 n}{2^n} \). We need to determine whether this sequence converges or diverges as \( n \to \infty \).

Step 2: Analyze the Denominator

Notice that the denominator \( 2^n \) grows exponentially as \( n \to \infty \). This means its value becomes very large very quickly.

Step 3: Evaluate the Numerator

The numerator is \( \cos^2 n \), which is a bounded oscillating function. Specifically, it satisfies \( 0 \leq \cos^2 n \leq 1 \) for all \( n \).

Step 4: Apply Limit Properties

To find the limit of \( a_n = \frac {\cos^2 n}{2^n} \), we use the property that if a bounded sequence is divided by an unbounded, increasingly large sequence, the result will tend toward 0.

Step 5: Calculate the Limit

Because \( \cos^2 n \) is bounded and \( 2^n \to \infty \), the sequence \( a_n = \frac {\cos^2 n}{2^n} \) converges to 0: \[ \lim_{{n \to \infty}} a_n = \lim_{{n \to \infty}} \frac {\cos^2 n}{2^n} = 0 \].

Key Concepts

Limit of a Sequence

When we speak about the limit of a sequence, we refer to the value that the terms of a sequence approach as the index goes to infinity. In mathematical terms, if we have a sequence \( \{a_n\} \), and \( \lim_{{n \to \infty}} a_n = L \), we say that the sequence converges to \( L \).
Finding limits is crucial in understanding convergence. For example, with the sequence \( a_n = \frac{\cos^2 n}{2^n} \), we want to understand what happens to \( a_n \) as \( n \) becomes very large.
By analyzing the components of \( a_n \), we recognize that although the numerator \( \cos^2 n \) oscillates between 0 and 1, the denominator \( 2^n \) grows very large due to exponential growth. This causes the fraction to become smaller and smaller, therefore approaching 0.

• The sequence has a limit if it settles down to a particular number.
• Understanding limits helps predict long-term behavior of the sequence.

In this exercise, we've determined that the limit is 0: \( \lim_{{n \to \infty}} a_n = 0 \). This conclusion is vital for establishing whether a sequence converges or diverges.

Bounded Functions

A bounded function is one that remains within a fixed range for all inputs. In our sequence, the function \( \cos^2 n \) is a perfect example. No matter what integer value \( n \) takes, \( \cos^2 n \) will always be between 0 and 1.
This characteristic of being bounded makes \( \cos^2 n \) easy to work with in convergence analysis. Why? Because we can confidently say that its maximum impact on the sequence is limited.

• Bounedness means stability, not wildly fluctuating.
• Provides assurance that \( \cos^2 n \) won’t excessively impact the convergence process.

In convergence discussions, a bounded numerator can help the sequence get closer to zero when divided by a rapidly growing denominator, much like how \( \frac{\cos^2 n}{2^n} \) behaves as \( n \) increases. This is why, despite the oscillation, the factor of being bounded aids in simplifying the determination of limits.

Exponential Growth

Exponential growth is a concept where quantities increase rapidly at a constant growth rate. In mathematics, this term often describes sequences or functions like \( 2^n \), as seen in our denominator.
This rapid increase grows so fast with each step that eventually dwarfs any bounded numerator like \( \cos^2 n \).
Let’s break it down:

• Imposes a quick expansion causing the entire fraction to shrink towards zero.
• Significantly affects convergence analysis because it determines the fraction's long-term behavior.

The exponential nature means the larger \( n \) gets, the much larger \( 2^n \) will be compared to \( \cos^2 n \), making the sequence's limit zero. Such growth rates make exponential functions valuable in mathematics for handling large-scale phenomena, which is why understanding them as part of sequences is crucial.