Maths encyclopedia and lessons  
Search

Mathematics Encyclopedia and Lessons

 
     
 

Lessons

Popular
Subjects

algebra
arithmetic
calculus
equations
geometry
differential equations
trigonometry
number theory
probability theory
more
 

References

applied mathematics
mathematical games
mathematicians
more
 
 

Alternating series

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

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

with an ≥ 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+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 the Cauchy criterion: if the sequence an 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 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.

01-04-2007 01:18:14
The contents of this article are licensed from Wikipedia.org
under the GNU Free Documentation License. How to see transparent copy