Kummer's transformation of series

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Short description: Mathematical method

In mathematics, specifically in the field of numerical analysis, Kummer's transformation of series is a method used to accelerate the convergence of an infinite series. The method was first suggested by Ernst Kummer in 1837.

Technique

Let

[math]\displaystyle{ A=\sum_{n=1}^\infty a_n }[/math]

be an infinite sum whose value we wish to compute, and let

[math]\displaystyle{ B=\sum_{n=1}^\infty b_n }[/math]

be an infinite sum with comparable terms whose value is known. If the limit

[math]\displaystyle{ \gamma:=\lim_{n\to \infty} \frac{a_n}{b_n} }[/math]

exists, then [math]\displaystyle{ a_n-\gamma \,b_n }[/math] is always also a sequence going to zero and the series given by the difference, [math]\displaystyle{ \sum_{n=1}^\infty (a_n-\gamma\, b_n) }[/math], converges. If [math]\displaystyle{ \gamma\neq 0 }[/math], this new series differs from the original [math]\displaystyle{ \sum_{n=1}^\infty a_n }[/math] and, under broad conditions, converges more rapidly.[1] We may then compute [math]\displaystyle{ A }[/math] as

[math]\displaystyle{ A=\gamma\,B + \sum_{n=1}^\infty (a_n-\gamma\,b_n) }[/math],

where [math]\displaystyle{ \gamma B }[/math] is a constant. Where [math]\displaystyle{ a_n\neq 0 }[/math], the terms can be written as the product [math]\displaystyle{ (1-\gamma\,b_n/a_n)\,a_n }[/math]. If [math]\displaystyle{ a_n\neq 0 }[/math] for all [math]\displaystyle{ n }[/math], the sum is over a component-wise product of two sequences going to zero,

[math]\displaystyle{ A=\gamma\,B + \sum_{n=1}^\infty (1-\gamma\,b_n/a_n)\,a_n }[/math].

Example

Consider the Leibniz formula for π:

[math]\displaystyle{ 1 \,-\, \frac{1}{3} \,+\, \frac{1}{5} \,-\, \frac{1}{7} \,+\, \frac{1}{9} \,-\, \cdots \,=\, \frac{\pi}{4}. }[/math]

We group terms in pairs as

[math]\displaystyle{ 1 - \left(\frac{1}{3} - \frac{1}{5}\right) - \left(\frac{1}{7} - \frac{1}{9}\right) + \cdots }[/math]
[math]\displaystyle{ \, = 1 - 2\left(\frac{1}{15} + \frac{1}{63} + \cdots \right) = 1-2A }[/math]

where we identify

[math]\displaystyle{ A = \sum_{n=1}^\infty \frac{1}{16n^2-1} }[/math].

We apply Kummer's method to accelerate [math]\displaystyle{ A }[/math], which will give an accelerated sum for computing [math]\displaystyle{ \pi=4-8A }[/math].

Let

[math]\displaystyle{ B = \sum_{n=1}^\infty \frac{1}{4n^2-1} = \frac{1}{3} + \frac{1}{15} + \cdots }[/math]
[math]\displaystyle{ \, = \frac{1}{2} - \frac{1}{6} + \frac{1}{6} - \frac{1}{10} + \cdots }[/math]

This is a telescoping series with sum value ​12. In this case

[math]\displaystyle{ \gamma := \lim_{n\to \infty} \frac{\frac{1}{16n^2-1}}{\frac{1}{4n^2-1}} = \lim_{n\to \infty} \frac{4n^2-1}{16n^2-1} = \frac{1}{4} }[/math]

and so Kummer's transformation formula above gives

[math]\displaystyle{ A=\frac{1}{4} \cdot \frac{1}{2} + \sum_{n=1}^\infty \left ( 1-\frac{1}{4} \frac{\frac{1}{4n^2-1}}{\frac{1}{16n^2-1}} \right ) \frac{1}{16n^2-1} }[/math]
[math]\displaystyle{ = \frac{1}{8} - \frac{3}{4} \sum_{n=1}^\infty \frac{1}{16n^2-1}\frac{1}{4n^2-1} }[/math]

which converges much faster than the original series.

Coming back to Leibniz formula, we obtain a representation of [math]\displaystyle{ \pi }[/math] that separates [math]\displaystyle{ 3 }[/math] and involves a fastly converging sum over just the squared even numbers [math]\displaystyle{ (2n)^2 }[/math],

[math]\displaystyle{ \pi=4-8A }[/math]
[math]\displaystyle{ =3+6\cdot\sum_{n=1}^\infty \frac{1}{(4(2n)^2-1)((2n)^2-1)} }[/math]
[math]\displaystyle{ =3 + \frac{2}{15} + \frac{2}{315} + \frac{6}{5005} + \cdots }[/math]

See also

  • Euler transform

References

  1. Holy et al., On Faster Convergent Infinite Series, Mathematica Slovaca, January 2008

External links