Absolute convergence and conditional convergence.

Definition 9.5.1   A series $\underset{n=0}{\overset{+\infty }{\sum}} u_n$ is absolutely convergent if the series $\underset{n=0}{\overset{+\infty }{\sum}} \vert u_n\vert$ is convergent.

Proposition 9.5.2   If a series is absolutely convergent, it is convergent.

Example 9.5.3   Consider the series $\underset{n=1}{\overset{+\infty }{\sum}} \frac {\cos n}{n^3}$ . The corresponding series of absolute values is $\underset{n=1}{\overset{+\infty }{\sum}} \left\vert \frac {\cos n}{n^3} \right\vert =\underset{n=1}{\overset{+\infty }{\sum}} \frac { \vert \cos n \vert }{n^3}$ is convergent because for any $n$ , its general term is less than or equal to $\frac {1}{n^3}$ (v.s. 3.3).

Thus, the given series is absolutely convergent, and is convergent.

For example, the series $\underset{n=0}{\overset{+\infty }{\sum}} \frac {(-1)^n}{n}$ is convergent, but not absolutely convergent, because the series $\underset{n=0}{\overset{+\infty }{\sum}} \frac 1n$ is divergent, by Prop. 3.1.

A convergent series which is convergent, but not absolutely convergent is conditionally convergent.

For example, the alternating harmonic series is convergent (v.s. 4.3), but not absolutely convergent (v.s. 1.9), thus it is conditionally convergent.

Theorem 9.5.4   Let $\underset{n=0}{\overset{+\infty }{\sum}} u_n$ and $\underset{n=0}{\overset{+\infty }{\sum}} v_n$ be two series, where the sequence $(v_n)$ is a rearrangement of the sequence $(u_n)$ .

If $\underset{n=0}{\overset{+\infty }{\sum}} u_n$ is absolutely convergent, then $\underset{n=0}{\overset{+\infty }{\sum}} v_n$ is absolutely convergent, and the two series have the same sum.

Remark 9.5.5   If the series $\underset{n=0}{\overset{+\infty }{\sum}} u_n$ is conditionally convergent, we can reaarange its terms so that it will show any predecided behaviour: divergence or convergence (with any limit we want!). For example, consider the alternating harmonic sequence.

• With the following rearrangement of terms

  $$\underset{n=1}{\overset{+\infty }{\sum}} \frac {(-1)^{n+1}}{2n-1} + \underset{n=1}{\overset{+\infty }{\sum}} \frac {(-1)^{n+1}}{2n}$$

  
  
  the series is divergent.
• Start with 1, substract $\frac 12$ , then add $\frac 13 + \frac 15$ , then add consecutive negative terms so that the sum will be less than 1, then add consecutive positive terms until the sum is 1 or more, and iterate this process. You get a convergent series whose sum is 1.