Kummer's transformation of series

Template:Short description 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

$A=\sum _{n=1}^{\mathrm{\infty }}{a}_n$

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

$B=\sum _{n=1}^{\mathrm{\infty }}{b}_n$

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

$\gamma :=\underset{n\to \mathrm{\infty }}{\lim }\frac{a_n}{b_n}$

exists, then $a_n-\gamma b_n$ is always also a sequence going to zero and the series given by the difference, $\sum _{n=1}^{\mathrm{\infty }}(a_n-\gamma b_n)$, converges. If $\gamma \ne 0$, this new series differs from the original $\sum _{n=1}^{\mathrm{\infty }}{a}_n$ and, under broad conditions, converges more rapidly. We may then compute $A$ as

$A=\gamma B+\sum _{n=1}^{\mathrm{\infty }}(a_n-\gamma b_n)$,

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

$A=\gamma B+\sum _{n=1}^{\mathrm{\infty }}(1-\gamma b_n/a_n)a_n$.

Example

Consider the Leibniz formula for π:

$1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\cdots =\frac{\pi }{4}.$

We group terms in pairs as

$1-(\frac{1}{3}-\frac{1}{5})-(\frac{1}{7}-\frac{1}{9})+\cdots$

$=1-2(\frac{1}{15}+\frac{1}{63}+\cdots )=1-2A$

where we identify

$A=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{16n^2-1}$.

We apply Kummer's method to accelerate $A$, which will give an accelerated sum for computing $\pi =4-8A$.

Let

$B=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{4n^2-1}=\frac{1}{3}+\frac{1}{15}+\cdots$

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

This is a telescoping series with sum value Template:Frac. In this case

$\gamma :=\underset{n\to \mathrm{\infty }}{\lim }\frac{\frac{1}{16n^2-1}}{\frac{1}{4n^2-1}}=\underset{n\to \mathrm{\infty }}{\lim }\frac{4n^2-1}{16n^2-1}=\frac{1}{4}$

and so Kummer's transformation formula above gives

$A=\frac{1}{4}\cdot \frac{1}{2}+\sum _{n=1}^{\mathrm{\infty }}(1-\frac{1}{4}\frac{\frac{1}{4n^2-1}}{\frac{1}{16n^2-1}})\frac{1}{16n^2-1}$

$=\frac{1}{8}-\frac{3}{4}\sum _{n=1}^{\mathrm{\infty }}\frac{1}{16n^2-1}\frac{1}{4n^2-1}$

which converges much faster than the original series.

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

$\pi =4-8A$

$=3+6\cdot \sum _{n=1}^{\mathrm{\infty }}\frac{1}{(4(2n{)}^2-1)((2n{)}^2-1)}$

$=3+\frac{2}{15}+\frac{2}{315}+\frac{6}{5005}+\cdots$

See also

• Euler transform

References

• Template:SpringerEOM
• Template:Cite book
• Template:Cite web
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External links

• Template:MathWorld

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