Problem 20 - Essential Calculus Early Transcendentals

Question

\(19-40=\) Determine whether the series is absolutely convergent, conditionally convergent, or divergent. $$ \sum_{n=1}^{\infty} \frac{(-2)^{n}}{n^{2}} $$

Step-by-Step Solution

Answer

The series \(\sum_{n=1}^{\infty} \frac{(-2)^n}{n^2}\) is divergent.

Step 1: Understand Absolute Convergence

A series \(\sum_{n=1}^{\infty} a_n\) is absolutely convergent if the series of absolute values \(\sum_{n=1}^{\infty} |a_n|\) converges. For the series given, we consider \(a_n = \frac{(-2)^n}{n^2}\). So, look at \(|a_n| = \frac{2^n}{n^2}\).

Step 2: Apply the Ratio Test to \(\sum_{n=1}^{\infty} \left| a_n \right|\)

Use the Ratio Test to determine if \(\sum_{n=1}^{\infty} \frac{2^n}{n^2}\) converges. Compute the limit:\[\lim_{n\to\infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n\to\infty} \frac{2^{n+1}/(n+1)^2}{2^n/n^2} = \lim_{n\to\infty} \frac{2 \cdot n^2}{(n+1)^2}\]Simplify:\[\lim_{n\to\infty} \frac{2n^2}{(n+1)^2} = \lim_{n\to\infty} \frac{2n^2}{n^2+2n+1}.\]

Step 3: Evaluate the Limit for Absolute Convergence Check

Divide the numerator and the denominator by \(n^2\):\[\lim_{n\to\infty} \frac{2n^2}{n^2+2n+1} = \lim_{n\to\infty} \frac{2}{1+\frac{2}{n}+\frac{1}{n^2}} = 2.\]Since this result (2) is greater than 1, by the Ratio Test, \(\sum_{n=1}^{\infty} \frac{2^n}{n^2}\) diverges. Therefore, \(\sum_{n=1}^{\infty} a_n\) is not absolutely convergent.

Step 4: Consider Conditional Convergence

A series \(\sum_{n=1}^{\infty} a_n\) is conditionally convergent if it converges, but the series of absolute values \(\sum_{n=1}^{\infty} |a_n|\) diverges. Apply the Alternating Series Test to \(\sum_{n=1}^{\infty} \frac{(-2)^n}{n^2}\).

Step 5: Apply the Alternating Series Test

For the alternating series \(\sum_{n=1}^{\infty} b_n\), where \(b_n = \frac{(-2)^n}{n^2}\), observe that the non-alternating term \(\frac{2^n}{n^2}\) does not decrease, and it does not satisfy the alternating series conditions, specifically, the terms do not approach zero as required for convergence.

Step 6: Conclude the Convergence

Since \(\sum_{n=1}^{\infty} |a_n|\) diverges and the alternating series test is not satisfied due to the growth of \(2^n\), the series \(\sum_{n=1}^{\infty} \frac{(-2)^n}{n^2}\) diverges. Hence, it is not absolutely convergent or conditionally convergent.

Key Concepts

Ratio Test

The ratio test is a powerful tool used to determine the convergence or divergence of an infinite series. It evaluates the "ratio" of consecutive terms in the series. If a series is given by \( \sum_{n=1}^{\infty} a_n \), the ratio test calculates the limit \( L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \). Here's what the outcomes mean:

• If \( L < 1 \), the series converges absolutely.
• If \( L > 1 \), the series diverges.
• If \( L = 1 \), the test is inconclusive.

For our series, \( \sum_{n=1}^{\infty} \frac{(-2)^n}{n^2} \), we first convert it using absolute values: \( \sum_{n=1}^{\infty} \frac{2^n}{n^2} \). Applying the ratio test here led to \( L = 2 \), which is greater than 1, indicating that this series diverges.

Alternating Series Test

The alternating series test applies specifically to series that alternate in sign, such as when a negative sign occurs in front of the term raised to the power of \(n\). For the alternating series \( \sum_{n=1}^{\infty} (-1)^n b_n \), the series converges if:

• \( b_{n+1} \leq b_n \) for all \( n \) (the terms are non-increasing).
• \( \, \lim_{n \to \infty} b_n = 0 \).

In our case, the series \( \sum_{n=1}^{\infty} \frac{(-2)^n}{n^2} \) does not fulfill these conditions. The term \( \frac{2^n}{n^2} \) grows, instead of diminishes, as \( n \) increases. Thus, the series fails the alternating series test, confirming that it does not converge.

Absolute Convergence

A series \( \sum_{n=1}^{\infty} a_n \) is said to be absolutely convergent if the series of absolute values \( \sum_{n=1}^{\infty} |a_n| \) converges. This is an important test because if a series is absolutely convergent, it is also convergent in the traditional sense.
In this exercise, we converted the series \( \sum_{n=1}^{\infty} \frac{(-2)^n}{n^2} \) into its absolute form: \( \sum_{n=1}^{\infty} \frac{2^n}{n^2} \). We then applied the ratio test to this new series and found that it diverged because the limit exceeded 1. Thus, the original series does not meet the criteria for absolute convergence.

Conditional Convergence

Conditional convergence occurs when a series \( \sum_{n=1}^{\infty} a_n \) converges, but the series \( \sum_{n=1}^{\infty} |a_n| \) diverges. This means that the series converges only because of the cancellation of alternating signs and not due to the magnitudes of the terms.
In our exercise, the presence of the negative factor and increasing \( 2^n \) meant that neither the series itself nor its absolute form converged. As a result, there's no conditional convergence here because the series must converge in the first place for conditional convergence to be considered. Since both the original and absolute series diverge, this series does not meet the definition of conditional convergence.