Kummer's transformation of series

We apply the method to accelerate the Leibniz formula for π:

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

First group terms in pairs as

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

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

where

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

Let

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

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

which is a telescoping series with sum ​¹⁄₂. In this case

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

and Kummer's transformation gives

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

This simplifies to

${\displaystyle A={\frac {1}{8}}-{\frac {3}{4}}\sum _{n=1}^{\infty }{\frac {1}{(16n^{2}-1)(4n^{2}-1)}}}$

which converges much faster than the original series.

• 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.
• Keith Conrad. "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.