Alternating series

In mathematics, an alternating series is an infinite series of the form

$\sum _{n=0}^{\infty }(-1)^{n}\,a_{n},$

with a_n ≥ 0. A sufficient condition for the series to converge is that it converges absolutely. But this is often too strong a condition to ask: it is not necessary. For example, the harmonic series

$\sum _{n=0}^{\infty }{\frac {1}{n}},$

diverges, while the alternating version

$\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{n}}$

converges to $\ln 2$.

A broader test for convergence of an alternating series is the Cauchy criterion: if the sequence $a_{n}$ is monotone decreasing and tends to zero, then the series

$\sum _{n=0}^{\infty }(-1)^{n}\,a_{n}$

converges.

A conditionally convergent series is an infinite series that converges, but does not converge absolutely. The following non-intuitive result is true: if the real series

$\sum _{n=0}^{\infty }(-1)^{n}\,a_{n}$

converges conditionally, then for every real number $\beta$ there is a reordering $\sigma$ of the series such that

$\sum _{n=0}^{\infty }(-1)^{n}\,a_{\sigma (n)}=\beta .$

As an example of this, consider the series above for $\ln 2$:

$\ln 2=\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{n}}=1-{\frac {1}{2}}+{\frac {1}{3}}-{\frac {1}{4}}+{\frac {1}{5}}-\cdots .$

One possible reordering for this series is as follows (the only purpose of the brackets in the first line is to help clarity):

$1-{\frac {1}{2}}-{\frac {1}{4}}+\left({\frac {1}{3}}-{\frac {1}{6}}\right)-{\frac {1}{8}}+\left({\frac {1}{5}}-{\frac {1}{10}}\right)-{\frac {1}{12}}+\left({\frac {1}{7}}-{\frac {1}{14}}\right)-{\frac {1}{16}}+\cdots$

$={\frac {1}{2}}-{\frac {1}{4}}+{\frac {1}{6}}-{\frac {1}{8}}+{\frac {1}{10}}-\cdots$

$={\frac {1}{2}}\left(1-{\frac {1}{2}}+{\frac {1}{3}}-{\frac {1}{4}}+{\frac {1}{5}}-\cdots \right)$

$={\frac {1}{2}}\,\ln 2.$