Question

Test the following series for convergence. . \(\sum_{\mathrm{n}=1}^{\infty} \frac{(-2)^{n}}{n^{2}}\)

Short answer

The series diverges.

Step-by-step solution

Identify the Series

Given series: \[ \sum_{n=1}^{\infty} \frac{(-2)^{n}}{n^{2}} \.\] This is an infinite series with general term \(a_n = \frac{(-2)^n}{n^2}\).

Apply the Absolute Convergence Test

To determine convergence, first check for absolute convergence by examining the series of the absolute values of the terms: \[\sum_{n=1}^{\infty} \left| \frac{(-2)^n}{n^2} \right| = \sum_{n=1}^{\infty} \frac{2^n}{n^2}\.\]

Use the Ratio Test

Apply the Ratio Test to the absolute series. Let \(a_n = \frac{2^n}{n^2}\) and evaluate\[\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \frac{2^{n+1}/(n+1)^2}{2^n/n^2} = \lim_{n \to \infty} \frac{2^{n+1} \cdot n^2}{2^n \cdot (n+1)^2} = \lim_{n \to \infty} \frac{2 \cdot n^2}{(n+1)^2}\.\]

Simplify the Limit

Simplify the limit:\[ \lim_{n \to \infty} \frac{2n^2}{(n+1)^2} = 2 \lim_{n \to \infty} \frac{n^2}{n^2 (1 + \frac{1}{n})^2} = 2 \lim_{n \to \infty} \frac{1}{(1 + \frac{1}{n})^2} = 2 \cdot \frac{1}{1^2} = 2\.\]

Interpret the Ratio Test Result

The Ratio Test result \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = 2 > 1 \) indicates that the series \( \sum_{n=1}^{\infty} \frac{2^n}{n^2} \) diverges. Therefore, the original series \( \sum_{n=1}^{\infty} \frac{(-2)^n}{n^2} \) does not converge absolutely.

Test for Conditional Convergence

Since the series does not converge absolutely, check for conditional convergence. For alternating series \( \sum (-1)^n b_n \) to converge, \( b_n \) must be positive, decreasing, and \( \lim_{n \to \infty} b_n = 0 \). However, the terms here are \( \frac{2^n}{n^2} \), which do not fit the criteria for an alternating series directly.

Conclusion

Since the series diverges by the Ratio Test and does not meet the criteria for conditional convergence, the series \( \sum_{n=1}^{\infty} \frac{(-2)^n}{n^2} \) diverges.

Key concepts

Absolute Convergence Test

When faced with a series, the first step to check for convergence is often to use the Absolute Convergence Test. This involves taking the absolute value of each term in the series and checking if this new series converges. For the given series, we have \(\frac{(-2)^n}{n^2} \).To check for absolute convergence, consider the absolute series \( \sum_{n=1}^{\infty} \left| \frac{(-2)^n}{n^2} \right| = \sum_{n=1}^{\infty} \frac{2^n}{n^2}\).If this absolute series converges, then the original series also converges absolutely.Absolute convergence is a very strong form of convergence. It implies that rearranging the terms of the series won’t affect the sum. This is immensely useful because it allows for more flexibility in manipulating the series.For our given series, determining that the absolute series diverges means we have to proceed to other tests to determine if the series might still conditionally converge or diverge completely.

Ratio Test

The Ratio Test is a popular method to determine convergence, especially when the series involves exponential terms. Here, for the series \(\sum_{n=1}^{\infty} \frac{2^n}{n^2}\),we apply the Ratio Test by taking the limit of the absolute value of the ratio of consecutive terms. Specifically, for \(a_n = \frac{2^n}{n^2}\),we compute:\[ \text{lim}_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \text{lim}_{n \to \infty} \frac{2^{n+1} \cdot n^2}{2^n \cdot (n+1)^2} \].Simplifying this results in:\[ \text{lim}_{n \to \infty} \frac{2 \cdot n^2}{(n+1)^2} = 2 \cdot \text{lim}_{n \to \infty} \frac{n^2}{(n+1)^2} = 2 \cdot \text{lim}_{n \to \infty} \frac{1}{(1+\frac{1}{n})^2} = 2 \cdot 1 = 2 \].Because the limit is greater than 1, the Ratio Test tells us that the series \(\sum_{n=1}^{\infty} \frac{2^n}{n^2}\) diverges. Since the absolute series diverges, this directly implies that the original series does not converge absolutely.

Divergence

After determining that a series does not converge absolutely, we need to consider if it converges conditionally. However, for the series in question, once we've determined that the absolute test fails and the terms don't meet the criteria for an alternating series test, we can conclude divergence.Conditional convergence requires that the terms of an alternating series decrease in magnitude and approach zero. In our series, \(\frac{2^n}{n^2}\), doesn't fit these requirements.Ultimately, because the ratio of the terms grows too rapidly (factor of 2), and does not balance out with the decay rate provided by the \(n^2\) term in the denominator, this implies the original series diverges.To summarize, applying the Ratio Test indicates divergence because the ratio exceeds 1. Also, the series fails to meet criteria for conditional convergence, affirming our conclusion that the series \(\sum_{n=1}^{\infty} \frac{(-2)^n}{n^2}\)diverges.