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
 * $$\gamma:=\lim_{n\to \infty} \frac{a_n}{b_n}$$

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. 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
 * $$1 - \left(\frac{1}{3} - \frac{1}{5}\right) - \left(\frac{1}{7} - \frac{1}{9}\right) + \cdots$$
 * $$\, = 1 - 2\left(\frac{1}{15} + \frac{1}{63} + \cdots \right) = 1-2A$$

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 $$(2n)^2$$,
 * $$\pi=4-8A$$
 * $$=3+6\cdot\sum_{n=1}^\infty \frac{1}{(4(2n)^2-1)((2n)^2-1)}$$
 * $$=3 + \frac{2}{15} + \frac{2}{315} + \frac{6}{5005} + \cdots$$