Alternating Harmonic Series is Conditionally Convergent

Theorem

$\ds \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^\paren {n - 1} } n = 1 - \frac 1 2 + \frac 1 3 - \frac 1 4 + \cdots$

Proof

Note first that:

$\ds \sum_{n \mathop = 1}^\infty \size {\frac {\paren {-1}^\paren {n - 1} } n} = \sum_{n \mathop = 1}^\infty \frac 1 n$

which is divergent by Harmonic Series is Divergent.

Next note that $\size {\dfrac 1 n}$ is a basic null sequence:

$\ds \lim_{n \mathop \to \infty} \dfrac 1 n = 0$

and that:

$\forall n \in \N_{>0}: \dfrac 1 n > 0$
$\dfrac 1 n > \dfrac 1 {n + 1}$

The result follows from the Alternating Series Test.

$\blacksquare$