## Absolute and Conditional Convergence |

An * alternating series* is a series of the form
\[\sum_{n=1}^{\infty} (-1)^n a_n=-a_1+a_2-a_3+a_4-\cdots \text{ or }
\sum_{n=1}^{\infty} (-1)^{n-1} a_n=a_1-a_2+a_3-a_4+\cdots,\]
where \(\{a_n\}\) is a sequence of positive real numbers.

**Example.**
The following alternating series is called the alternating harmonic series
\[\displaystyle \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots.\]

** Alternating Series Test:**
The alternating series \(\displaystyle\sum_{n=1}^{\infty} (-1)^n a_n\) and \(\displaystyle\sum_{n=1}^{\infty} (-1)^{n-1} a_n\)
are convergent if \(\{a_n\}\) is a positive and decreasing sequence that converges to \(0\), i.e.,

- \(a_n>0\) for all integers \(n\geq 1\),
- \(a_{n+1} < a_n\) for all integers \(n\geq 1\), and
- \(\displaystyle\lim_{n\to \infty} a_n=0\).

Consider the alternating series \(\displaystyle\sum_{n=1}^{\infty} (-1)^n a_n\) where \(\{a_n\}\) is a positive
and decreasing sequence that converges to \(0\). Since \(\{a_n\}\) is positive and decreasing,
\[s_{2n}=s_{2n-2}+(-a_{2n-1}+a_{2n}) < s_{2n-2}.\]
Thus \(\{s_{2n}\}\) is a decreasing sequence. Note that
\[s_{2n}=-a_1+(a_2-a_3)+\cdots+(a_{2n-2}-a_{2n-1})+a_{2n}.\]
Since \((a_2-a_3),\ldots,(a_{2n-2}-a_{2n-1}),a_{2n}\) are positive, \(s_{2n} > -a_1\). Thus \(\{s_{2n}\}\) is a
decreasing sequence that is bounded below. Then by the Monotone Convergence Theorem, \(\{s_{2n}\}\) is convergent.
Suppose \(\displaystyle\lim_{n\to \infty} s_{2n}=s\). Since \(s_{2n+1}=s_{2n}-a_{2n+1}\),
\[\lim_{n\to \infty} s_{2n+1}=\lim_{n\to \infty} s_{2n}- \lim_{n\to \infty} a_{2n+1}=s-0=s.\]
Since both \(\{s_{2n}\}\) and \(\{s_{2n+1}\}\) converge to \(s\), so does \(\{s_{n}\}\). Therefore
\[\sum_{n=1}^{\infty} (-1)^n a_n=\lim_{n\to \infty} s_{n}=s.\]

**Example.**
Use the Alternating Series Test to show that \(\displaystyle \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}\)
is convergent.

* Solution.*
The given series is \(\displaystyle\sum_{n=1}^{\infty} (-1)^{n-1} a_n\) where \(a_n=\frac{1}{n}\).
\(\{a_n\}=\{\frac{1}{n}\}\) is a positive and decreasing sequence that converges to \(0\) because

- \(a_n=\frac{1}{n} > 0\) for all integers \(n\geq 1\),
- \(a_{n+1}=\frac{1}{n+1} < \frac{1}{n}=a_n\) for all integers \(n\geq 1\), and
- \(\displaystyle\lim_{n\to \infty} a_n =\lim_{n\to \infty}\frac{1}{n}=0\).

Thus by the Alternating Series Test, \(\displaystyle \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}\) is convergent.

Note:

- To show that \(\{a_n\}\) is decreasing where the formula of \(a_n=f(n)\) is more complicated, show that
\(f'(x) < 0\) on \((1,\infty)\). For the above problem, \(f(x)=\frac{1}{x}\) and \(f'(x)=-\frac{1}{x^2} < 0\)
on \( (1,\infty)\). Then \( \{a_n\}=\{\frac{1}{n}\}\) is decreasing.
- In the above example, \(\displaystyle \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}\) is convergent but \(\displaystyle \sum_{n=1}^{\infty} \left\vert \frac{(-1)^{n-1}}{n}\right\vert =\sum_{n=1}^{\infty} \frac{1}{n}\) is divergent.

**Definition.** A series \(\displaystyle \sum_{n=1}^{\infty} a_n\) is * absolutely convergent* if
\(\displaystyle \sum_{n=1}^{\infty} |a_n|\) is convergent. A series \(\displaystyle \sum_{n=1}^{\infty} a_n\)
is * conditionally convergent* if it is convergent but \(\displaystyle \sum_{n=1}^{\infty} |a_n|\) is
divergent (i.e., a conditionally convergent series is a convergent series that is not absolutely convergent).

**Example.**
Recall that \(\displaystyle \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}\) is convergent but
\(\displaystyle \sum_{n=1}^{\infty} \left\vert \frac{(-1)^{n-1}}{n}\right\vert =\sum_{n=1}^{\infty} \frac{1}{n}\)
is divergent. Therefore \(\displaystyle \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}\) is conditionally convergent
but not absolutely convergent.

**Theorem.**
An absolutely convergent series is convergent, i.e., if \(\displaystyle \sum_{n=1}^{\infty} |a_n|\) is convergent,
then \(\displaystyle \sum_{n=1}^{\infty} a_n\) is also convergent.

Suppose that \(\displaystyle \sum_{n=1}^{\infty} |a_n|\) is convergent. For all integers \(n\geq 1\),
\(-|a_n|\leq a_n \leq |a_n|\). Adding \(|a_n|\), we get
\[0\leq a_n+|a_n|\leq 2|a_n|.\]
Since \(\displaystyle \sum_{n=1}^{\infty} |a_n|\) is convergent, so is \(\displaystyle \sum_{n=1}^{\infty} 2|a_n|=2\sum_{n=1}^{\infty} |a_n|\).
By the Comparison Test, \(\displaystyle \sum_{n=1}^{\infty} (a_n+|a_n|)\) is convergent.
Finally \(\displaystyle \sum_{n=1}^{\infty} a_n\) is also convergent as the difference of two convergent series:
\[ \sum_{n=1}^{\infty} a_n= \sum_{n=1}^{\infty} (a_n+|a_n|) -\sum_{n=1}^{\infty} |a_n|.\]

**Example.**
Consider the series \(\displaystyle \sum_{n=1}^{\infty} \frac{\sin n}{n^2}\) which has positive and negative
terms. For all integers \(n\geq 1\),
\[ \left\vert\frac{\sin n}{n^2}\right\vert\leq \frac{1}{n^2}. \]
Since \(\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^2}\) is convergent by the \(p\)-series Test with \(p=2 > 1\),
\(\displaystyle \sum_{n=1}^{\infty} \left\vert\frac{\sin n}{n^2}\right\vert\) is convergent by the Comparison Test.
Since \(\displaystyle \sum_{n=1}^{\infty} \left\vert\frac{\sin n}{n^2}\right\vert\) is convergent, \(\displaystyle \sum_{n=1}^{\infty} \frac{\sin n}{n^2}\)
is absolutely convergent and consequently convergent.

To see the importance of the concept of absolute and conditional convergence of a series, we discuss rearrangements of the terms of a series. Consider the alternating harmonic series \(\displaystyle \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}\) whose sum is obtained by evaluating the Taylor series of \(\ln(1+x)\) at \(x=1\). \[\displaystyle \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots =\ln 2.\] Now we rearrange the terms as follows. \[\begin{align*} &\;\;\left( 1-\frac{1}{2} \right)-\frac{1}{4}+ \left( \frac{1}{3}-\frac{1}{6} \right)-\frac{1}{8}+\cdots\\ &=\frac{1}{2}-\frac{1}{4}+\frac{1}{6}-\frac{1}{8}+\cdots\\ &=\frac{1}{2}\left( 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots \right)\\ &=\frac{1}{2}\ln 2. \end{align*}\] We can even rearrange the terms to make it converge to any real number. This is due to the conditional convergence of the alternating harmonic series.

** Riemann's Rearrangement Theorem:**
Let \(\displaystyle \sum_{n=1}^{\infty} a_n\) be a conditionally convergent series and \(L\) be a real number.
Then there is a rearrangement of \(\displaystyle \sum_{n=1}^{\infty} a_n\) that converges to \(L\).

The above strange phenomenon does not happen for absolutely convergent series.

**Theorem.**
Let \(\displaystyle \sum_{n=1}^{\infty} a_n\) be an absolutely convergent series converging to \(L\).
Then any rearrangement of \(\displaystyle \sum_{n=1}^{\infty} a_n\) also converges to \(L\).

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