Problem 39 - Calculus Early Transcendentals: Pearson New International Edition

Question

Show that \(\lim _{n \rightarrow \infty} a_{n}=0\) is not sufficient to guarantee the convergence of the alternating series \(\Sigma(-1)^{n+1} a_{n} .\) Hint: Alternate the terms of \(\sum 1 / n\) and \(\Sigma\left(-1 / n^{2}\right)\).

Step-by-Step Solution

Answer

The condition \( \lim_{n \rightarrow \infty} a_n = 0 \) is not sufficient for convergence without the series being absolutely convergent or other criteria met.

Step 1: Understand Alternating Series Convergence

An alternating series of the form \( \Sigma (-1)^{n+1} a_n \) converges if the sequence \( a_n \) is decreasing and its limit as \( n \rightarrow \infty \) is zero. We will investigate if \( \lim_{n \rightarrow \infty} a_n = 0 \) alone (without the sequence being decreasing) is enough for convergence.

Step 2: Examine Series \( \Sigma \frac{1}{n} \)

The harmonic series \( \Sigma \frac{1}{n} \) is known to diverge. The sequence \( a_n = \frac{1}{n} \) satisfies \( \lim_{n \rightarrow \infty} a_n = 0 \), but this condition alone does not make it converge.

Step 3: Construct Alternating Series with Harmonic Terms

Consider the alternating series \( S = \Sigma (-1)^{n+1} \frac{1}{n} \). This alternates terms from the divergent harmonic series and satisfies \( \lim_{n \rightarrow \infty} \frac{1}{n} = 0 \). We check if it converges.

Step 4: Test Alternating Series for Convergence

For the series \( S \) to converge, the terms \( \frac{1}{n} \) should be monotonically decreasing, which they are. However, the convergence of its absolute series \( \Sigma \frac{1}{n} \) is not satisfied (since the harmonic series diverges), indicating \( S \) potentially diverges or doesn't meet typical convergence tests fully.

Step 5: Examine Alternative Series \( \Sigma \left(-\frac{1}{n^2}\right) \)

Now consider and construct an alternating series with \( a_n = \frac{1}{n^2} \), \( T = \Sigma (-1)^{n+1} \frac{1}{n^2} \). This new series converges absolutely since \( \Sigma \frac{1}{n^2} \) converges (p-series with \( p > 1 \)).

Step 6: Analyze Impact of Convergence Criteria

Combining findings, the convergence of an alternating series is not assured by \( \lim_{n \rightarrow \infty} a_n = 0 \) alone, illustrated by \( S \) potentially not converging as \( \Sigma \frac{1}{n} \) diverges, while \( T \) converges due to a stricter criterion.

Key Concepts

Harmonic Series

The harmonic series is a well-known series in mathematics, defined as \[ \sum_{n=1}^{\infty} \frac{1}{n} . \]This series is particularly interesting because it seems to suggest it might converge since each individual term \( \frac{1}{n} \) gets smaller as \( n \) increases. However, the harmonic series is actually divergent. This means that as you add more and more terms, the sum doesn't approach a fixed number, but instead increases without bound.

• The harmonic series demonstrates a critical point in the study of series: just because the terms \( a_n \) tend to zero, it does not guarantee that the series itself converges.
• This characteristic is essential for understanding why the convergence of alternating series requires more than just having terms diminish to zero.

Convergence Criteria

Understanding the convergence of a series is an integral part of series analysis. For an alternating series, which takes the general form \[ \Sigma (-1)^{n+1} a_n , \]we have specific criteria to determine its convergence. The alternating series test stipulates that an alternating series converges if two conditions are met:

• The absolute value of the terms \( a_n \) decreases monotonically, meaning each term is smaller than the ones before it.
• The limit of the sequence of terms \( a_n \) as \( n \rightarrow \infty \) is zero.

This means that even if \( a_n \) approaches zero, that's not enough on its own. The behavior of the entire series must be considered. For instance, revisiting the harmonic series, if we apply its terms in an alternating series, it still might not converge because the original series itself diverges.

Absolute Convergence

Absolute convergence is a stronger form of convergence for series. A series \[ \Sigma a_n \]is said to be absolutely convergent if the series formed by taking the absolute values of its terms \[ \Sigma |a_n| \]converges. This is a powerful property because if a series converges absolutely, it also converges normally, without the alternating sign conditions required by the alternating series test.

• An example of absolute convergence can be seen in the series \( \Sigma \frac{1}{n^2} \), which definitely converges because it is a p-series with \( p = 2 \) (where \( p > 1 \), ensuring convergence).
• When series converge absolutely, rearranging their terms won't affect their sum, which isn't always the case with conditionally convergent series.

In the case of the alternating series formed by \( \left(-\frac{1}{n^2}\right) \), the absolute series \( \Sigma \frac{1}{n^2} \) converging confirms its absolute convergence, providing a contrast to harmonic series's lack of convergence.