Acceleration of convergence of slow series

It’s fun (to me too) to accelarate classical slow series like

$\log(2)\ =\ \frac 11 - \frac 12 + \frac 13 - \ldots$

$\frac{\pi}4\ =\ \frac 11 - \frac 13 + \frac 15 - \ldots$

The latter is called Gregory-Leibniz series. What about the former? Is it Gregory-Leibniz too? Anyway, I played with them in the past and have reproduced, after Euler, versions which converge almost as fast as linear combinations of Taylor series. Possibly my web pages are still somewhere buried deep quietly.

Let $a_n\ (n=0\ 1\ \ldots)$ be an arithmetic progression of positive real numbers. Let $d := a_1-a_2$ be the difference of this progression. Finally, let $k$ be a positive integer. Then

$\sum_{n=0}^\infty\ \frac 1{\prod_{j=0}^k a_{n+j}}\ \ =\ \ \frac1{k\cdot d}\cdot\frac1{\prod_{j=0}^{k-1} a_{n+j}}$

This allows to accelerate similar series by approximating them by the ones based on arithmetic series as above. A given series gets split into one of the above and the difference, which is a new series. But the new series converges faster. Euler was able to iterate this simple acceleration operation infinitely many times in a way which resulted in fast converging series.

This time I’d like to look at this method a bit more systematically than in the past. Instead of selecting the above easily summable auxiliary series artistically I’d like to do it perhaps optimally (or to make a compromise but a conscientious compromise).
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Let me give an idea of what I am talking about on the initial examples. Thus first let’s transform the above G-L series::

$\log(2)\ \ =\ \ \sum_{n=0}^\infty\ \frac 1{(2\cdot n+1)\cdot(2\cdot n+2)}$

$\frac{\pi}4\ \ =\ \ \sum_{n=0}^\infty\ \frac 1{(4\cdot n+1)\cdot(4\cdot n+3)}\ \ =\ \ 2\cdot\sum_{n=1}^\infty\ \frac 1{(4\cdot n-1)\cdot(4\cdot n+1)}$

Now

$\frac 1{(2\cdot n+1)\cdot(2\cdot n+2)}\ \ =\ \ \frac 1 {(2\cdot n+\frac 12)\cdot(2\cdot n+\frac 52)}\,\ -\,\ \frac 34\cdot\frac 1{(2\cdot n+\frac 12)\cdot(2\cdot n+1)\cdot(2\cdot n+2)\cdot(2\cdot n+\frac 52)}$

hence

$\log(2)\ \ =\ \ 1\ -\ 3\cdot\sum_{n=0}^\infty \frac 1{(2\cdot n+1)\cdot(2\cdot n+2)\cdot(4\cdot n+1)\cdot(4\cdot n+5)}$

Blastog

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Acceleration of convergence of slow series

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