Absolute convergence of series

## Absolute convergence of series
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            ## Absolute convergence
            #### Definition
            If $\sum\_{n=1}^\infty \lvert a_n \rvert$ converges, then the series $\sum\_{n=1}^\infty a_n$ is said to be *absolutely convergent*.
            #### Theorem
            If, a series is absolutely convergent, then it is convergent
            #### Example
            Because $\sum\_{n=1}^\infty \frac1{n^3}$ converges, $\sum\_{n=1}^\infty \frac{(-1)^n}{n^3}$ is absolutely convergent, hence also converges.

#### Example
            Which of

$\sum\_{n=1}^\infty \frac{(-1)^n}{n},\\; \sum\_{n=1}^\infty \frac{n+3}{2-3n-2n^2},\\; \sum\_{n=1}^\infty \frac{500\sin(\pi n/2)}{n^2},\\; \sum\_{n=1}^\infty \frac{(-1)^n\times3}{n-5\pi},\\; \sum\_{n=1}^\infty \frac{(-1)^nn^2}{n+100}$

can you tell are are absolutely convergent, convergent, or divergent?
            
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            ## Absolute convergence
            #### Definition
            If $\sum\_{n=1}^\infty \lvert a_n \rvert$ converges, then the series $\sum\_{n=1}^\infty a_n$ is said to be *absolutely convergent*.
            #### Theorem
            If, a series is absolutely convergent, then it is convergent
            #### Example
            Because $\sum\_{n=1}^\infty \frac1{n^3}$ converges, $\sum\_{n=1}^\infty \frac{(-1)^n}{n^3}$ is absolutely convergent, hence also converges.

#### Example
            The series $\sum\_{n=1}^\infty \frac{(-1)\^{n+1}}{n}$ is not absolutely convergent, but it is convergent.

A series which is convergent but not absolutely convergent is called *conditionally convergent*.
            A conditionally convergent series will converge to different values if you reorder its terms.
            
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            ## Conditional convergence
            #### Example
            The series $S = 1-1+\frac12-\frac12+\frac13-\frac13+\frac14-\frac14+\ldots$ is not absolutely convergent, but it is convergent.

Indeed $S=0$.

But, if we could reorder the terms without changing $S$, then take two positive terms before each negative term to get

$S = 1+\frac12-1+\frac13+\frac14-\frac12+\ldots$

$S = 1-\frac12+\frac13-\frac14+\ldots = \log(2) = \sum\_{n=1}^\infty \frac{(-1)\^{n+1}}{n}$, as we shall prove later.