Problem 13 - Calculus: Early Transcendentals

Question

Determine whether the series converges or diverges. \( \displaystyle \sum_{n= 1}^{\infty} \frac {1 + \cos n}{e^n} \)

Step-by-Step Solution

Answer

The series converges.

Step 1: Identify the Series and Terms

We are given the infinite series \( \sum_{n=1}^{\infty} \frac{1 + \cos n}{e^n} \). Each term of the series is \( a_n = \frac{1 + \cos n}{e^n} \).

Step 2: Determine a Suitable Test for Convergence

Consider using comparison tests or the Limit Comparison Test, as \( e^n \) grows rapidly and might dominate the behavior of terms well.

Step 3: Basic Comparison - Exponential Series

Notice that \( |1 + \cos n| \leq 2 \) since \( \cos n \) oscillates between \(-1\) and \(1\). So, \( 0 \leq \frac{1 + \cos n}{e^n} \leq \frac{2}{e^n} \).

Step 4: Convergence of a Simplified Exponential Series

The series \( \sum_{n=1}^{\infty} \frac{2}{e^n} \) is a geometric series with a common ratio \( r = \frac{1}{e} < 1 \), which converges.

Step 5: Apply the Squeeze Theorem

Since \( 0 \leq \frac{1 + \cos n}{e^n} \leq \frac{2}{e^n} \) and \( \sum_{n=1}^{\infty} \frac{2}{e^n} \) converges, by the Squeeze Theorem, \( \sum_{n=1}^{\infty} \frac{1 + \cos n}{e^n} \) also converges.

Key Concepts

Comparison Test

The Comparison Test is a powerful tool in determining series convergence. It involves comparing the series in question to a second series that is known to converge or diverge.
If the terms of your series are smaller than the terms of a known convergent series, your series also converges. On the flip side, if the terms are larger than those of a divergent series, it diverges.
Here’s how you use it:

• Identify your series, in this case, \( \sum_{n=1}^{\infty} \frac{1 + \cos n}{e^n} \).
• Find a series \( \sum b_n \) for comparison. Commonly, a simpler geometric or p-series is used.
• Ensure \( 0 \leq a_n \leq b_n \). If \( \sum b_n \) converges and \( a_n \leq b_n \), then \( \sum a_n \) converges by the test.

In this exercise, \( \frac{1 + \cos n}{e^n} \leq \frac{2}{e^n} \) fits perfectly due to properties of \( \cos n \). Success in comparison shows likely convergence!

Geometric Series

Geometric series are simple yet fascinating. They take the form\( \sum_{n=0}^{\infty} ar^n \), where \( a \) is the initial term and \( r \) is the common ratio.
A geometric series converges if \( |r| \lt 1 \), and it diverges if \( |r| \geq 1 \).
To determine this, just focus on the ratio:

• For a convergent series, sum is \( \frac{a}{1-r} \).
• For a divergent series, it grows infinitely.

This principle helps simplify complex expressions. In our solution, \( \sum_{n=1}^{\infty} \frac{2}{e^n} \) is recognized as a geometric series given \( a = 2,\) and \( r = \frac{1}{e} \), meeting our convergence criterion \( \frac{1}{e} < 1 \).
This relation becomes a critical part in using the Squeeze Theorem!

Squeeze Theorem

The Squeeze Theorem is a nuanced method for proving convergence of series. It works by 'squeezing' a series between two other series. If both outer series converge to the same limit, then our target series must converge to that limit too.
Here's the breakdown:

• Identify \( a_n \), \( b_n \), and \( c_n \).
• Ensure \( a_n \leq c_n \leq b_n \).
• Prove \( \lim_{n \to \infty} a_n = \lim_{n \to \infty} b_n \).

If both limits exist and are equal, then \( \lim_{n \to \infty} c_n \) exists and equals them.
In this exercise, we fit the series \( \sum_{n=1}^{\infty} \frac{1 + \cos n}{e^n} \) into \( \sum_{n=1}^{\infty} 0 \) and \( \sum_{n=1}^{\infty} \frac{2}{e^n} \). Both bounds pointed to convergence, proving our series converges.

Convergence of Exponential Series

Exponential series converge based on growing denominators. As the exponential grows faster, terms shrink quickly, aiding convergence.
Look for decreasing exponential functions, such as \( e^n \), in your series denominator.
In mathematical terms:

• If \( a_n = \frac{f(n)}{e^{kn}} \), where \( k > 0 \), observe \( f(n) \) and compare with recognizable series.
• Decreasing terms denote convergence—proven using other tests like the Comparison Test.

In the exercise, \( \frac{1 + \cos n}{e^n} \) shows exponential decay due to \( e^n \), leading us through a rational comparison with decreasing geometric expressions.
Efficient convergence assessment makes exponential series approachable and leads to accurate conclusions.