Alternating series

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

\sum_{n=0}^\infty (-1)^n\,a_n,

with an ≥ 0. A finite sum of this kind is an alternating sum.

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+1},

diverges, while the alternating version

\sum_{n=0}^\infty \frac{(-1)^n}{n+1}

converges to the natural logarithm of 2.

A broader test for convergence of an alternating series is Leibniz' test: if the sequence an is monotone decreasing and tends to zero, then the series

\sum_{n=0}^\infty (-1)^n\,a_n

converges.

The partial sum

s_n = \sum_{k=0}^n (-1)^k a_k

can be used to approximate the sum of a convergent alternating series. If an is monotone decreasing and tends to zero, then the error in this approximation is less than an + 1.

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 β there is a reordering σ of the series such that

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

As an example of this, consider the series above for the natural logarithm of 2:

\ln 2=\sum_{n=0}^\infty \frac{(-1)^n}{n+1}=1-\frac12+\frac13-\frac14+\frac15-\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-\frac12-\frac14+\left(\frac13-\frac16\right)-\frac18+\left(\frac15-\frac1{10}\right)-\frac1{12} +\left(\frac17-\frac1{14}\right)-\frac1{16}+\cdots
=\frac12-\frac14+\frac16-\frac18+\frac1{10}-\cdots
=\frac12\left(1-\frac12+\frac13-\frac14+\frac15-\cdots\right)
=\frac12\,\ln2.

A proof of this assertion runs along the lines: the greedy algorithm for σ is correct.

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