Absolute and Conditional Convergence

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The basic question we wish to answer about a series is whether or not the series converges. If a series has both positive and negative terms, we can refine this question and ask whether or not the series converges when all terms are replaced by their absolute values. This is the distinction between absolute and conditional convergence, which we explore in this section.

Recall that the basic question about a series that we seek to answer is “does the series converge?” It turns out that if the series contains negative terms, there is an interesting refinement of this question. This is illustrated by the following example.

We have seen that the alternating harmonic series

\[ \sum _{n=1}^\infty \frac {(-1)^n}{n} \]

converges. On the other hand, if we construct a new series by taking the absolute value of each term, we obtain

\[ \sum _{n=1}^\infty \left | \frac {(-1)^n}{n} \right | = \sum _{n=1}^\infty \frac {1}{n}. \]

That is, we obtain the standard harmonic series, which is one of our favorite examples of a divergent series.

This example shows that it is interesting to consider the role that negative terms play in the convergence of a series.

A series $\sum a_n$ converges absolutely if $\sum |a_n|$ converges.
A series $\sum a_n$ converges conditionally if $\sum a_n$ converges but $\sum |a_n|$ diverges.

We then refine the basic question about a series (“does the series converge or diverge?”) to the following, more subtle, question: “does the series converge absolutely, converge conditionally, or diverge?”

Does the series

\[ \sum _{n=1}^\infty (-1)^n\frac {n+3}{n^2+2n+5} \]

converge absolutely, converge conditionally, or diverge?

The series converges conditionally. The series converges absolutely. The series diverges.

We can show the series

\[ \sum _{n=1}^\infty \left |(-1)^n\frac {n+3}{n^2+2n+5}\right |= \sum _{n=1}^\infty \frac {n+3}{n^2+2n+5} \]

diverges using the limit comparison test, comparing with $1/n$. The series

\[ \sum _{n=1}^\infty (-1)^n\frac {n+3}{n^2+2n+5} \]

converges using the alternating series test; we conclude the series converges conditionally.

Does the series

\[ \sum _{n=1}^\infty (-1)^n\frac {n^2+2n+5}{2^n} \]

converge absolutely, converge conditionally, or diverge?

The series converges conditionally. The series converges absolutely. The series diverges.

We can show the series

\[ \sum _{n=1}^\infty \left |(-1)^n\frac {n^2+2n+5}{2^n}\right |=\sum _{n=1}^\infty \frac {n^2+2n+5}{2^n} \]

converges using the ratio test. Therefore we conclude

\[ \sum _{n=1}^\infty (-1)^n\frac {n^2+2n+5}{2^n} \]

converges absolutely.

By definition, a series converges conditionally when $\sum a_n$ converges but $\sum |a_n|$ diverges. Conversely, one could ask whether it is possible for $\sum |a_n|$ to converge while $\sum a_n$ diverges. The following theorem shows that this is not possible.

Absolute Convergence Theorem Every absolutely convergent series must converge.

Said differently, if a series $\sum |a_n|$ converges, then the series $\sum a_n$ must also converge. It is not hard to see why this is true. The terms of any sequence $\{a_n\}$ (possibly containing negative terms) satisfy the inequalities

\[ 0 \leq a_n + |a_n| \leq 2|a_n|. \]

If we assume that $\sum |a_n|$ converges, then $\sum (a_n + |a_n|)$ must also converge by the Comparison Test. But then the series $\sum a_n$ converges as well, as it is the difference of a pair of convergent series:

\[ \sum a_n = \sum (a_n + |a_n|) - \sum |a_n|. \]

Does the series $\sum _{n=1}^\infty \frac {\sin (n)}{n^2}$ converge?

The series contains both positive and negative terms, but it is not alternating. This makes it difficult to apply our standard tests to determine whether the series converges directly. On the other hand, consider the series

\[ \sum _{n=1}^\infty \left |\frac {\sin (n)}{n^2}\right |. \]

By design, all of its terms are nonnegative. Moreover, since $|\sin (n)| \leq 1$, we have the comparison

\[ \left |\frac {\sin (n)}{n^2} \right | \leq \frac {1}{n^2}. \]

It follows by the Comparison Test that $\sum \left |\frac {\sin (n)}{n^2}\right |$ converges. We conclude that $\sum \frac {\sin (n)}{n^2}$ converges absolutely, and the Absolute Convergence Theorem implies that it must therefore converge.

2025-01-06 18:32:32