Absolutely Convergent Series
Given a series \(\sum_{n=1}^\infty a_n\), consider the corresponding series \(\sum_{n=1}^\infty\lvert a_n\rvert=\lvert a_1\rvert+\lvert a_2\rvert+\lvert a_3\rvert+....\). The series \(\sum_{n=1}^\infty a_n\) is absolutely convergent if the series of absolute values \(\sum_{n=1}^\infty|a_n|\) is convergent.
Theorem: If a series is absolutely convergent, then it is convergent. Note that the reverse is not true (the series in Example (a) below is convergent, but not absolutely convergent).
Conditionally Convergent Series
A series \(\sum_{n=1}^\infty a_n\) is conditionally convergent if it is convergent, but not absolutely convergent.
Example
Determine if each series is absolutely convergent or conditionally convergent.
(a) \(\sum_{n=1}^\infty\frac{(-1)^{n-1}}n\)
(b) \(\sum_{n=1}^\infty\frac{(-1)^{n-1}}{3^n}\)
Solution
(a) \(\sum_{n=1}^\infty\lvert a_n\rvert=\sum_{n=1}^{\infty}\left|\frac{(-1)^{n-1}}{n}\right|=\sum_{n=1}^\infty\frac1n\). This is a p-series with \(p=1\), and thus, diverges.
But, \(\sum_{n=1}^\infty\frac{(-1)^{n-1}}n\) converges by the Alternating Series Test. Thus, \(\sum_{n=1}^\infty\frac{(-1)^{n-1}}n\) is conditionally convergent.
(b) \(\sum_{n=1}^{\infty}|a_n|=\sum_{n=1}^{\infty}\left|\frac{(-1)^{n-1}}{3^n}\right|=\sum_{n=1}^{\infty}\frac{1}{3^n}\). This is a geometric series with \(r=\frac{1}{3}<1\) which converges. Thus, \(\sum_{n=1}^\infty\frac{(-1)^{n-1}}{3^n}\) is absolutely convergent (and thus, convergent as well).