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.

Technique

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

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

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

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

Example

Consider the Leibniz formula for π: $${\displaystyle 1\,-\,{\frac {1}{3}}\,+\,{\frac {1}{5}}\,-\,{\frac {1}{7}}\,+\,{\frac {1}{9}}\,-\,\cdots \,=\,{\frac {\pi }{4}}.}$$ We group terms in pairs as $${\displaystyle {\begin{aligned}&1-\left({\frac {1}{3}}-{\frac {1}{5}}\right)-\left({\frac {1}{7}}-{\frac {1}{9}}\right)+\cdots \\&\quad =1-2\left({\frac {1}{15}}+{\frac {1}{63}}+\cdots \right)=1-2A\end{aligned}}}$$ where we identify $${\displaystyle A=\sum _{n=1}^{\infty }{\frac {1}{16n^{2}-1}}.}$$ We apply Kummer's method to accelerate ${\displaystyle A}$, which will give an accelerated sum for computing ${\displaystyle \pi =4-8A}$.

Let $${\displaystyle {\begin{aligned}B&=\sum _{n=1}^{\infty }{\frac {1}{4n^{2}-1}}={\frac {1}{3}}+{\frac {1}{15}}+\cdots \\&={\frac {1}{2}}-{\frac {1}{6}}+{\frac {1}{6}}-{\frac {1}{10}}+\cdots \end{aligned}}}$$ 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}}} and so Kummer's transformation formula above gives $${\displaystyle {\begin{aligned}A&={\frac {1}{4}}\cdot {\frac {1}{2}}+\sum _{n=1}^{\infty }\left(1-{\frac {1}{4}}{\frac {\frac {1}{4n^{2}-1}}{\frac {1}{16n^{2}-1}}}\right){\frac {1}{16n^{2}-1}}\\&={\frac {1}{8}}-{\frac {3}{4}}\sum _{n=1}^{\infty }{\frac {1}{16n^{2}-1}}{\frac {1}{4n^{2}-1}}\end{aligned}}}$$ which converges much faster than the original series.

Coming back to Leibniz formula, we obtain a representation of ${\displaystyle \pi }$ that separates ${\displaystyle 3}$ and involves a fastly converging sum over just the squared even numbers ${\displaystyle (2n)^{2}}$, $${\displaystyle {\begin{aligned}\pi &=4-8A\\&=3+6\cdot \sum _{n=1}^{\infty }{\frac {1}{(4(2n)^{2}-1)((2n)^{2}-1)}}\\&=3+{\frac {2}{15}}+{\frac {2}{315}}+{\frac {6}{5005}}+\cdots \end{aligned}}}$$

See also

• Euler transform

References

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

• 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.

External links

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