THC Science

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[edit]

Let

A=\sum _{n=1}^{\infty }a_{n}

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

B=\sum _{n=1}^{\infty }b_{n}

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

\gamma :=\lim _{n\to \infty }{\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, $\sum _{n=1}^{\infty }(a_{n}-\gamma \,b_{n})$ , converges. If $\gamma \neq 0$ , this new series differs from the original $\sum _{n=1}^{\infty }a_{n}$ and, under broad conditions, converges more rapidly.^[1] We may then compute $A$ as

A=\gamma \,B+\sum _{n=1}^{\infty }(a_{n}-\gamma \,b_{n})

,

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

A=\gamma \,B+\sum _{n=1}^{\infty }(1-\gamma \,b_{n}/a_{n})\,a_{n}

.

Example[edit]

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

\,=1-2\left({\frac {1}{15}}+{\frac {1}{63}}+\cdots \right)=1-2A

where we identify

A=\sum _{n=1}^{\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}^{\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 1⁄2. In this case

\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}}

and so Kummer's transformation formula above gives

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}}

={\frac {1}{8}}-{\frac {3}{4}}\sum _{n=1}^{\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 $(2n)^{2}$ ,