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 $A=\sum _{n=1}^{\infty }a_{n}$ 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 ${\displaystyle \gamma$ exists, then $a_{n}-\gamma \,b_{n}$ 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 $\gamma \neq 0$ , 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 $A$ 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

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

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

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