Section 6.6 Absolute and Conditional Convergence
¶Roughly speaking there are two ways for a series to converge: As in the case of \(\sum 1/n^2\text{,}\) the individual terms get small very quickly, so that the sum of all of them stays finite, or, as in the case of \(\ds \sum (1)^{n1}/n\text{,}\) the terms don't get small fast enough (\(\sum 1/n\) diverges), but a mixture of positive and negative terms provides enough cancellation to keep the sum finite. You might guess from what we've seen that if the terms get small fast enough that the sum of their absolute values converges, then the series will still converge regardless of which terms are actually positive or negative. This leads us to the following theorem.
Theorem 6.54. Absolute Convergence Test.
If \(\ds\sum_{n=0}^\infty a_n\) converges, then \(\ds\sum_{n=0}^\infty a_n\) converges.
Proof.
Note that \(\ds 0\le a_n+a_n\le 2a_n\) so by the Comparison Test \(\ds\sum_{n=0}^\infty (a_n+a_n)\) converges. Since
by Theorem 6.29, the conclusion follows.
Note:
Intuitively, this theorem says that it is (potentially) easier for \(\sum a_n\) to converge than for \(\sum a_n\) to converge, because terms may partially cancel in the first series.

So given a series \(\sum a_n\) with both positive and negative terms, you should first ask whether \(\sum a_n\) converges. This may be an easier question to answer, because we have tests that apply specifically to series with nonnegative terms.
If \(\sum a_n\) converges then you know that \(\sum a_n\) converges as well.
If \(\sum a_n\) diverges then it still may be true that \(\sum a_n\) converges, but you will need to use other techniques to decide.
If \(\sum a_n\) converges we say that \(\sum a_n\) converges absolutely. To say that \(\sum a_n\) converges absolutely is to say that the terms of the series get small (in absolute value) quickly enough to guarantee that the series converges, regardless of whether any of the terms cancel each other. For example \(\ds\sum_{n=1}^\infty (1)^{n1} {1\over n^2}\) converges absolutely.
Definition 6.55. Absolutely Convergent.
Given a series \(\ds\sum_{n=1}^{\infty} a_n\text{.}\) If the corresponding series \(\ds\sum_{n=1}^{\infty} a_n\) converges, then \(\ds\sum_{n=1}^{\infty} a_n\) converges absolutely.
If \(\sum a_n\) converges but \(\sum a_n\) does not, we say that \(\sum a_n\) converges conditionally. Recall that the alternating harmonic series \(\ds\sum_{n=1}^{\infty} \frac{(1)^{n1}}{n}\) converges, but that the corresponding series of absolute values, namely the harmonic series \(\ds\sum_{n=1}^{\infty}\frac{1}{n}\text{,}\) diverges. Hence, the alternating harmonic series is conditionally convergent.
Definition 6.56. Conditionally Convergent.
Given a series \(\ds\sum_{n=1}^{\infty} a_n\text{.}\) If \(\ds\sum_{n=1}^{\infty} a_n\) converges, but the corresponding series \(\ds\sum_{n=1}^{\infty} a_n\) does not converge, then \(\ds\sum_{n=1}^{\infty} a_n\) converges conditionally.
Example 6.57. Absolutely Convergent, Conditionally Convergent, or Divergent.
Does \(\ds\sum_{n=2}^\infty {\sin n\over n^2}\) converge? If so, how?
In Example 6.53 we saw that \(\ds\sum_{n=2}^\infty {\sin n\over n^2}\) converges, so the given series converges absolutely.
Example 6.58. Absolutely Convergent, Conditionally Convergent, or Divergent.
Does \(\ds\sum_{n=0}^\infty (1)^{n}{3n+4\over 2n^2+3n+5}\) converge? If so, how?
We begin by analyzing the corresponding series of absolute values, where
Looking at the leading terms in the numerator and denominator of \(a_n\text{,}\) we speculate to compare this series with the harmonic series:
Indeed, \(n^2+n > 5\) for \(n >1\text{,}\) and therefore this series diverges since the harmonic series diverges. So if the original series converges, it does so conditionally.
To test for convergence, we apply the Alternating Series Test:
Clearly, \(a_n = \dfrac{3n+4}{2n^2+3n+5}\) is positive for all \(n \geq 0\text{.}\)
\(\lim\limits_{n\to\infty} \dfrac{3n+4}{2n^2+3n+5} = 0\text{.}\)
If we let \(f(x)= \dfrac{3x+4}{2x^2+3x+5}\) then \(f'(x) = \dfrac{6x^2+16x3}{(2x^2+3x+5)^2}\text{,}\) and it is not hard to see that this is negative for \(x \geq 1\text{,}\) so the sequence \(\left\{\frac{3n+4}{2n^2+3n+5}\right\}_{n=0}^{\infty}\) is decreasing.
Thus, the series \(\ds\sum_{n=0}^\infty (1)^{n}{3n+4\over 2n^2+3n+5}\) converges by the Alternating Series Test, and we conclude that the series converges conditionally.
Exercises for Section 6.6.
Exercise 6.6.1.
Determine whether each series converges absolutely, converges conditionally, or diverges.

\(\ds\sum_{n=1}^\infty (1)^{n1}{1\over 2n^2+3n+5}\)
AnswerSolutionconverges absolutelyWe first notice that the series\begin{equation*} \sum_{n=1}^{\infty} \frac{1}{2n^2+3n+5} \end{equation*}converges by comparison to the convergent \(p\)series \(\sum \frac{1}{n^2}\text{.}\) Next, we check that\begin{equation*} a_n = \frac{1}{2n^2+3n+5} > 0, \end{equation*}and\begin{equation*} \lim_{n\to\infty} a_n = 0, \end{equation*}and\begin{equation*} a_{n+1} \leq a_n \implies 2(n+1)^2 + 3(n+1) + 5 \geq 2n^2 + 3n +5, \end{equation*}which is true. Therefore, the series converges absolutely. 
\(\ds\sum_{n=1}^\infty (1)^{n1}{3n^2+4\over 2n^2+3n+5}\)
AnswerSolutiondivergesSince\begin{equation*} \begin{split}\lim_{n\to\infty} a_n \amp = \lim_{n\to\infty} \frac{3n^2+4}{2n^2+3n+5} \\ \amp \Heq \lim_{n\to\infty} \frac{6n}{4n+3}\\ \amp \Heq \lim_{n\to\infty} \frac{6}{4} \neq 0,\end{split} \end{equation*}by the \(n\)th Term Test, the series diverges. 
\(\ds\sum_{n=1}^\infty (1)^{n1}{\ln n\over n}\)
AnswerSolutionconverges conditionallyFor \(n\geq 4\text{,}\) \(\ln n > 1\) and so
\begin{equation*} \frac{\ln n}{n} > \frac{1}{n}\text{.} \end{equation*}Thus the series
\begin{equation*} \sum_{n=0}^{\infty} \frac{\ln n}{n} \end{equation*}diverges by comparison to the harmonic series. To determine whether the original series converges conditionally or diverges, we apply the Alternating Series Test: Let
\begin{equation*} a_n = \frac{\ln n}{n}, \ \ \text{ and } \ \ f(x) = \frac{\ln x}{x}\text{.} \end{equation*}Then
\begin{equation*} f'(x) = \diff{}{x} \frac{\ln x}{x} = \frac{1\ln x}{x^2}\text{,} \end{equation*}and so \(f'(x) \lt 0\) for \(x > e\text{.}\) Thus the sequence \(\{a_n\}_{n=1}^{\infty}\) is decreasing for \(n \geq 4\text{.}\) Additionally, we see that
\begin{equation*} \lim_{n\to\infty} a_n = \lim_{n\to\infty} \frac{\ln n}{n} = \lim_{n\to\infty} \frac{1/n}{1} = 0\text{.} \end{equation*}Therefore,
\begin{equation*} \lim_{n\to\infty} \frac{\ln n}{n} \end{equation*}is conditionally convergent.

\(\ds\sum_{n=1}^\infty (1)^{n1} {\ln n\over n^3}\)
AnswerSolutionconverges absolutelyWe first notice that\begin{equation*} a_n = \frac{\ln(n)}{n^3} \geq 0, \end{equation*}and\begin{equation*} \lim_{n\to\infty} a_n \Heq \lim_{n\to\infty} \frac{1}{3n^3} = 0, \end{equation*}and\begin{equation*} \diff{}{x} \frac{\ln(x)}{x} = \frac{13\ln(x)}{x^4} \lt 0 \quad \text{for } x\in (e^{1/3}, \infty). \end{equation*}Hence, by the Alternating Series Test, the series converges. Furthermore, the corresponding series\begin{equation*} \sum_{n=1}^{\infty} \frac{\ln (n)}{n^3} \end{equation*}converges by comparison with the convergent \(p\)series \(\sum \frac{1}{n^2}\text{,}\) since \(\ln (n) \lt n \text{.}\) Therefore, the alternating series converges absolutely. 
\(\ds\sum_{n=2}^\infty (1)^n{1\over \ln n}\)
AnswerSolutionconverges conditionallyWe first analyze the corresponding series of absolute values,
\begin{equation*} \sum_{n=2}^{\infty} \frac{1}{\ln n}\text{.} \end{equation*}For \(n \geq 2\text{,}\) we have that
\begin{equation*} \frac{1}{\ln n} > \frac{1}{n}\text{,} \end{equation*}and so the series diverges (since the harmonic series is divergent). Therefore, the original series either diverges or is conditionally convergent. We apply the Alternating Series Test:
\begin{equation*} \lim_{n\to\infty} \frac{1}{\ln n} = 0\text{,} \end{equation*}and for \(n \geq 2\text{,}\)
\begin{equation*} \ln (n+1) > \ln n \implies \frac{1}{\ln (n+1)} \lt \frac{1}{\ln n} \end{equation*}and also \(\ln n \gt 0\text{.}\) Thus, the series
\begin{equation*} \sum_{n=2}^{\infty} \frac{(1)^n}{\ln n} \end{equation*}is conditionally convergent.

\(\ds\sum_{n=0}^\infty (1)^{n} {3^n\over 2^n+5^n}\)
AnswerSolutionconverges absolutelyWe first notice that the corresponding series\begin{equation*} \sum_{n=1}^{\infty} \frac{3^n}{2^n+5^n} \end{equation*}converges by comparison with the convergent geometric series \(\sum \left(\frac{3}{5}\right)^n\text{.}\) We now check the conditions for the Alternating Series Test:\begin{equation*} a_n = \frac{3^n}{2^n+5^n} \geq 0, \end{equation*}and\begin{equation*} \frac{3^{n+1}}{2^n+5^n} \leq \frac{2^{n+1} + 5^{n+1}}{3^n} \implies a_{n+1} \leq a_n, \end{equation*}and\begin{equation*} \lim_{n\to\infty} a_n = \lim_{n\to\infty} \frac{3^n}{2^n+5^n} = 0. \end{equation*}Hence, the series converges absolutely. 
\(\ds\sum_{n=0}^\infty (1)^{n} {3^n\over 2^n+3^n}\)
AnswerSolutiondivergesWe have that\begin{equation*} \lim_{n\to\infty} \frac{3^n}{2^n+3^n} = \lim_{n\to\infty} \frac{1}{\left(\frac{2}{3}\right)^n+1} = 1, \end{equation*}and so by the \(n\)th Term Test, the series diverges. 
\(\ds\sum_{n=1}^\infty (1)^{n1} {\arctan n\over n}\)
AnswerSolutionconverges conditionallyWe first notice that the corresponding series\begin{equation*} \sum_{n=1}^{\infty} \frac{\arctan(n)}{n} \end{equation*}diverges since\begin{equation*} \frac{\arctan(n)}{n} \geq \frac{1}{n} \end{equation*}for all \(n \gt \tan(1).\) Therefore, if the given series converges, it does so conditionally. We now use the Alternating Series Test:\begin{equation*} a_n = \frac{\arctan(n)}{n} \geq 0, \end{equation*}and\begin{equation*} \lim_{n\to\infty} \frac{\arctan(n)}{n} \Heq \lim_{n\to\infty} \frac{1/(n^2+1)}{1}= 0. \end{equation*}Now let \(f(x) = \frac{\arctan(x)}{x}\text{.}\) Since\begin{equation*} f'(x) = \frac{x\arctan(x) (x^2+1)}{x^2(x^2+1)} \lt 0 \end{equation*}for all \(x \gt 0\text{,}\) the corresponding sequence \(\{a_n\}\) is decreasing. Therefore, the series converges conditionally.