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There have been various questions on this site about whether there is any upper limit to how fast a series sum of rational numbers can converge on another rational. With the right choice of particular series there appears to be no upper limit.

The most common example more generally is the infinite geometric series

$$\frac{1}{1-z}=z^0+z^1+z^2+...$$

However a much faster converging series is $$\frac{1}{z}=\sum_{k=1}^\infty \frac{1}{(k+z)\prod_{n=0}^{k-2}(n+z)}\tag{1}$$ which can be proved by the telescoping of the partial fraction

$$\frac{1}{z}=\frac{1}{z\left(1+z \right)}+\frac{1}{\left(1+z \right)}$$

These two series can be combined to give the double series $$\frac{z}{z-1}=\sum_{m=0}^\infty \sum_{k=1}^\infty \frac{1}{(k+z^m)\prod_{n=0}^{k-2}(n+z^m)}$$

Are there known examples of generally applicable infinite series for any given rational, that converge faster than (1)?

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  • $\begingroup$ For equation (1) what is the domain of convergence for $z$? $\endgroup$ – Somos Nov 19 '17 at 22:42
  • $\begingroup$ @Somos: As its just an telescoped partial fraction ultimately any $z$ for which $\frac{1}{z}$ is defined will converge. Not just rationals, but any finite real or complex number ($z=a+bi$) excepting $z=0$ will converge. $\endgroup$ – James Arathoon Nov 20 '17 at 0:11
  • $\begingroup$ What if z is a negative integer? What if $z=-1$? $\endgroup$ – Somos Nov 20 '17 at 3:31
  • $\begingroup$ @Somos: Your right because there are denominator factors with $(1+z)$, $(2+z)$ etc in sequential terms the series sum is undefined for $z=0$ or any negative integer. $\endgroup$ – James Arathoon Nov 20 '17 at 10:16
  • $\begingroup$ @Somos: Analogous to the Gamma Function is that what you are getting at. $\endgroup$ – James Arathoon Nov 20 '17 at 10:23

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