Kummer's transformation of series

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.

Let

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

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

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

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

${\displaystyle \gamma :=\lim _{n\to \infty }{\frac {a_{n}}{b_{n}}}}${\displaystyle \gamma :=\lim _{n\to \infty }{\frac {a_{n}}{b_{n}}}}

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

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

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

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

Consider the Leibniz formula for π:

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

We group terms in pairs as

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

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

where we identify

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

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

Let

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

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

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

and so Kummer's transformation formula above gives

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

${\displaystyle ={\frac {1}{8}}-{\frac {3}{4}}\sum _{n=1}^{\infty }{\frac {1}{16n^{2}-1}}{\frac {1}{4n^{2}-1}}}${\displaystyle ={\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 ${\displaystyle \pi }${\displaystyle \pi } that separates ${\displaystyle 3}${\displaystyle 3} and involves a fastly converging sum over just the squared even numbers ${\displaystyle (2n)^{2}}${\displaystyle (2n)^{2}},

${\displaystyle \pi =4-8A}${\displaystyle \pi =4-8A}

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

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

• Euler transform

• Senatov, V.V. (2001) [1994], "Kummer transformation", Encyclopedia of Mathematics , EMS Press
• Knopp, Konrad (2013). Theory and Application of Infinite Series. Courier Corporation. p. 247. ISBN 9780486318615.
• Conrad, Keith. "Accelerating Convergence of Series" (PDF).
• Kummer, E. (1837). "Eine neue Methode, die numerischen Summen langsam convergirender Reihen zu berech-nen". J. Reine Angew. Math. (16): 206–214.

• Weisstein, Eric W. "Kummer's Series Transformation". MathWorld .

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