Problem 27 - Calculus Early Transcendentals: Pearson New International Edition

Question

\(\sum_{n=1}^{\infty} \frac{1}{2+\sin ^{2} n}\)

Step-by-Step Solution

Answer

The series \( \sum_{n=1}^{\infty} \frac{1}{2+\sin^{2} n} \) diverges.

Step 1: Understanding the Series

We are given the infinite series \( \sum_{n=1}^{\infty} \frac{1}{2+\sin ^{2} n} \). Notice that the denominator is \(2+\sin^2 n\), which is always greater than 2 since \(\sin^2 n\) is non-negative.

Step 2: Analyzing Term Value

For each term in the series, \(\frac{1}{2+\sin^2 n}\), the denominator is always greater than 2. Therefore, the value of each term is less than \(\frac{1}{2}\).

Step 3: Applying the Direct Comparison Test

The series \( \sum \frac{1}{2+\sin^2 n} \) can be compared to the series \( \sum \frac{1}{2} \). Notice that each term of the original series is less than \( \frac{1}{2} \). Since the series \( \sum \frac{1}{2} \) is divergent (according to the Harmonic series), this comparison suggests that the original series is divergent.

Key Concepts

Direct Comparison Test

The Direct Comparison Test is a valuable tool when determining the convergence or divergence of an infinite series. It involves comparing the series in question with another series that has known behavior. To apply this test, you should:

• Identify a series with known convergence or divergence properties that is similar to the series you are examining.
• Ensure each term of your series is less than or greater than the corresponding term of the comparison series.

If a series with smaller terms than a known divergent series diverges, so does your series. Conversely, if a series with larger terms than a known convergent series converges, then your series must also converge. In the original solution, the series \( \sum \frac{1}{2+\sin^2 n} \) uses this test against the series \( \sum \frac{1}{2} \) because each term \( \frac{1}{2+\sin^2 n} \) is less than \( \frac{1}{2} \). Since \( \sum \frac{1}{2} \) is divergent, the Direct Comparison Test indicates divergence here too.

Convergence and Divergence

Convergence and divergence are core concepts when dealing with infinite series. Understanding these terms is crucial for analyzing series as they describe whether a series approaches a finite limit (converges) or does not (diverges).

• A series converges if the sum of its infinite terms approaches a specific finite value.
• A series diverges if the sum of its terms grows indefinitely or does not approach a finite value.

Various tests exist like the Direct Comparison Test, Ratio Test, and Root Test to analyze convergence or divergence. In our example, the series cannot converge because it is compared to an already divergent series of greater or equal terms. Identifying convergence or divergence helps in determining whether the sum of the infinite terms results in a meaningful value or not.

Harmonic Series

The harmonic series is one of the most famous and simplest examples of a divergent series. Despite each of its terms decreasing and converging toward zero, the series as a whole grows without bound.The general form of a harmonic series is:\[ \sum_{n=1}^{\infty} \frac{1}{n} \]This series is divergent, meaning it doesn’t sum to a finite value. It's important not to confuse the decreasing behavior of the terms with the behavior of the entire series. In the comparison from the original exercise, an altered harmonic-like series, \( \sum \frac{1}{2} \), was used to demonstrate that the original series diverges. Understanding the nature of harmonic series can be helpful for learning how seemingly shrinking series can still result in divergence.