Problem 20 - Thomas Calculus

Question

Which of the series converge, and which diverge? Use any method, and give reasons for your answers. \begin{equation}\sum_{n=1}^{\infty} \frac{1+\cos n}{n^{2}}\end{equation}

Step-by-Step Solution

Answer

The series converges by the Comparison Test.

Step 1: Understanding the Series

The series given is \( \sum_{n=1}^{\infty} \frac{1+\cos n}{n^{2}} \). It features the cosine function and a denominator consisting of \( n^2 \). Our first step is to figure out whether the cosine function could affect the convergence.

Step 2: Comparison with Known Series

Consider the corresponding series without the cosine term, \( \sum_{n=1}^{\infty} \frac{1}{n^{2}} \). This is a p-series with \( p = 2 \), which is known to converge.

Step 3: Establishing Bounds

Since \( -1 \leq \cos n \leq 1 \), we have \( 0 \leq 1+\cos n \leq 2 \). Therefore, \( 0 \leq \frac{1+\cos n}{n^2} \leq \frac{2}{n^2} \). The series \( \sum \frac{2}{n^2} \) converges as it is simply twice a convergent p-series.

Step 4: Applying the Comparison Test

Using the Comparison Test, because \( 0 \leq \frac{1+\cos n}{n^2} \leq \frac{2}{n^2} \), and \( \sum \frac{2}{n^2} \) converges, by comparison, \( \sum \frac{1+\cos n}{n^2} \) also converges.

Key Concepts

Comparison Test

The Comparison Test is a useful tool in determining the convergence or divergence of positive term series. This test works by comparing the terms of a given series with those of a second, known series.
To use the Comparison Test, we consider two series: a complex series we're interested in and a simpler reference series whose behavior (either convergence or divergence) we understand well. For the series given:

• If the terms of the complex series are bounded above by those of a convergent reference series, then the complex series also converges.
• If, conversely, the terms of the complex series are bounded below by those of a divergent series, then it too diverges.

In the original exercise, the series \( \sum \frac{1+\cos n}{n^2} \) was tested using the Comparison Test against the known convergent p-series \( \sum \frac{1}{n^2} \). Because each term of the series with the cosine function was shown to be less than or equal to \( \frac{2}{n^2} \), and since \( \sum \frac{2}{n^2} \) converges, the original series converges as well.

p-series

A p-series is a specific type of series of the form \( \sum \frac{1}{n^p} \), where \( p \) is a positive constant. The convergence or divergence of a p-series depends on the value of \( p \):

• If \( p > 1 \), the p-series converges.
• If \( p \leq 1 \), the p-series diverges.

In the step-by-step solution, the series \( \sum \frac{1}{n^2} \) served as a reference p-series for comparison. Here, \( p = 2 \), which satisfies the condition for convergence (since 2 is greater than 1). Using such a known convergent series helps in demonstrating the convergence of a more complex series, such as one involving other functions like the cosine function.

Cosine Function

The cosine function, denoted as \( \cos(n) \), is a periodic function that oscillates between -1 and 1. For series that include cosine terms, understanding its behavior is crucial.

• The periodic nature of cosine means it does not grow indefinitely and thus does not influence the overall growth of terms in a way that strongly affects convergence on its own.
• In convergence tests, as shown in the original exercise, the values of \( \cos(n) \) are used to establish bounds. Here, the inequality \( -1 \leq \cos n \leq 1 \) was utilized to transform the original series to an equivalent series bounded between 0 and \( 2/n^2 \).

This helps maintain the comparison of terms, ensuring that the convergence behavior observed in simple series can be applied to more complex series involving the cosine function.