Question

Convergence of Alternating Series For each of the following alternating series, determine whether the series converges or diverges. a. \(\sum_{n=1}^{\infty}(-1)^{n+1} / n^{2}\) b. \(\sum_{n=1}^{\infty}(-1)^{n+1} n /(n+1)\)

Short answer

a) Converges; b) Diverges.

Step-by-step solution

Identify the Series and Apply Alternating Series Test for Part a

For part a, we recognize the series:\[\sum_{n=1}^{\infty} (-1)^{n+1} \frac{1}{n^2}\]To apply the Alternating Series Test, check the following conditions:1. The absolute value of the terms \(a_n = \frac{1}{n^2}\) should be decreasing.2. \(\lim_{n \to \infty} a_n = 0\).Since \(\frac{1}{n^2}\) is decreasing and tends to 0 as \(n\) approaches infinity, the series converges by the Alternating Series Test.

Identify the Series and Check Convergence for Part b

For part b, the series is:\[\sum_{n=1}^{\infty} (-1)^{n+1} \frac{n}{n+1}\]To apply the Alternating Series Test, evaluate:1. Check if the sequence \(a_n = \frac{n}{n+1}\) decreases.2. Verify \(\lim_{n \to \infty} a_n = 0\).Here, \(a_n = \frac{n}{n+1}\) does not decrease for all \(n\), and \(\lim_{n \to \infty} a_n = 1 eq 0\). Therefore, the series diverges.

Key concepts

Convergence

In mathematics, convergence is a key concept when dealing with infinite series. Essentially, a series converges if the sum of its terms approaches a specific finite limit. When we say that an infinite series converges, it means adding more terms in the series will get you closer but not exceed a certain number.
This is important because not all series have this property. For some, as we add more terms, the sum continually grows without bound, indicating divergence. Understanding the behavior of an infinite series helps in determining whether it converges or diverges.
For the alternating series like \(\sum_{n=1}^{\infty}(-1)^{n+1} / n^{2}\), we often use specific tests like the Alternating Series Test to establish this convergent behavior. This test helps confirm that despite the oscillating nature of the series, the partial sums approach a definite value.

Divergence

Divergence is the opposite of convergence. A series diverges if its terms, when summed, never settle down to limit but rather keep growing.
Consider the series \(\sum_{n=1}^{\infty}(-1)^{n+1} n /(n+1)\). In this case, applying tests like the Alternating Series Test reveals that the series diverges. The reason is that the sequence \(a_n =\frac{n}{n+1}\)\, does not meet the criteria required for convergence.
Specifically, the terms of the series don't tend to zero, which is a crucial factor for convergence in alternating series. Recognizing divergence is essential in mathematics, as it can signal that we need a different approach or indicate the unbounded nature of the series under consideration.
Diverging series, unlike their converging counterparts, resist settling down to any specific value, defying predictability as terms are summed.

Alternating Series Test

The Alternating Series Test is a handy tool in determining the convergence of series with alternating signs, typically indicated by terms like \((-1)^{n+1}\)\ or similar expressions. It states that an alternating series \(\sum (-1)^{n}a_{n}\)\ converges if two conditions are met:

• The absolute value of the terms, \(a_n\)\, is decreasing for each subsequent term.
• The limit of the absolute values of the terms as \(n\rightarrow\infty\)\, is zero, i.e., \(\lim_{n\to\infty}a_n = 0\).

The test ensures that, despite their alternating nature, these series can still converge, given these conditions, making it a powerful investigative tool in calculus.
For example, in the series \(\sum_{n=1}^{\infty} (-1)^{n+1} \frac{1}{n^2}\)\, we apply the Alternating Series Test. Here, we confirm both conditions are satisfied since each term \(\,\frac{1}{n^2}\)\, is positive, decreasing, and tends toward zero as \(n\approaches\infty\).\ Hence, the series converges.

Limits

Limits are fundamental to understanding the behavior of functions and series, particularly as the index of a sequence grows towards infinity. In the context of series, assess the limit of the sequence of terms to determine convergence or divergence.
If the limit of a series' terms does not equal zero, the series cannot converge. This is demonstrated in part b of our original exercise, where the series \(\sum_{n=1}^{\infty} (-1)^{n+1} \frac{n}{n+1}\)\ diverges because \(\lim_{n\to\infty} \frac{n}{n+1}=1 eq 0\).
Limits give rise to precision in these assessments, offering a numeric boundary that terms are approaching, which is crucial for making accurate convergence assessments. Understanding and calculating limits is a skill equipping you to tackle more advanced calculus problems with confidence.