I remembered, and managed to find, still gathering dust in a forgotten corner of the Internet, an old QuickBASIC program which, with a trick, can rapidly sum up a HUGE amount of terms of the famous Madhava-Gregory-Leibniz series (MGL, or just "Gregory's series") for pi:
$$\frac{\pi}{4} = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9} - \cdots$$
The old thing is located here:
http://www.boo.net/~jasonp/suprgreg.bas
and the comments at the beginning of the program claim that the following series can be found by some kind of "transformation" of the MGL series:
$$\frac{\pi}{4} = \frac{5}{6} - \frac{11}{14} \frac{1 \cdot 2}{3 \cdot 5 \cdot 2} + \frac{17}{22} \frac{1 \cdot 2}{3 \cdot 5 \cdot 2} \frac{3 \cdot 4}{7 \cdot 9 \cdot 2} - \frac{23}{30} \frac{1 \cdot 2}{3 \cdot 5 \cdot 2} \frac{3 \cdot 4}{7 \cdot 9 \cdot 2} \frac{5 \cdot 6}{11 \cdot 13 \cdot 2} + \cdots$$
Where on earth does that come from?!
EDIT: I used to say here that the formula did not converge correctly. It does -- I made a mistake when entering in the fractions to the calculation program to test the formula and misled myself.
Assuming the pattern continues, this formula looks to be
$$\frac{\pi}{4} = \sum_{n=0}^{\infty} (-1)^n \frac{5 + 6n}{6 + 8n} \prod_{j=0}^{n-1} \frac{(2j+1)(2j+2)}{2(4j+3)(4j+5)}$$
Does it ring a bell?
Let's use technique like Euler's transform:
$$ \dfrac{\pi}{4} = 1-\dfrac{1}{3}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{13}-\dfrac{1}{15} + \cdots $$
$ =a_0-a_1+a_2-a_3+a_4-a_5+a_6-a_7+\cdots \\ = a_0 - \dfrac{1}{2}a_1 - \dfrac{1}{2}(a_1-a_2) + \dfrac{1}{2}(a_2-a_3) - \dfrac{1}{2}(a_3-a_4) + \dfrac{1}{2}(a_4-a_5) - \cdots \\ = \color{red}{\dfrac{5}{2\cdot 3}} - \dfrac{1}{3 \cdot 5} + \dfrac{1}{5\cdot 7} - \dfrac{1}{7 \cdot 9} + \dfrac{1}{9\cdot 11} - \cdots $
$~$
$ = \color{red}{\dfrac{5}{2\cdot 3}} - b_1 + b_2 - b_3+b_4-b_5+b_6-b_7+ \cdots \\ = \color{red}{\dfrac{5}{2\cdot 3}} - b_1 + \dfrac{1}{2}b_2 + \dfrac{1}{2}(b_2-b_3) - \dfrac{1}{2}(b_3-b_4) + \dfrac{1}{2}(b_4-b_5) - \cdots \\ = \color{red}{\dfrac{5}{2\cdot 3} - \dfrac{11}{2\cdot 3 \cdot 5 \cdot 7}} + \dfrac{2!}{5 \cdot 7 \cdot 9} - \dfrac{2!}{7 \cdot 9 \cdot 11} + \dfrac{2!}{9 \cdot 11 \cdot 13} - \cdots $
$~$
$ = \color{red}{\dfrac{5}{2\times 3} - \dfrac{11}{2\times 3 \cdot 5 \cdot 7}} + c_2 - c_3 + c_4 - c_5 + c_6 - c_7 + \cdots \\ = \color{red}{\dfrac{5}{2\times 3} - \dfrac{11}{2\times 3 \cdot 5 \cdot 7}} + c_2 - \dfrac{1}{2}c_3 -\dfrac{1}{2}(c_3-c_4) +\dfrac{1}{2}(c_4-c_5) -\dfrac{1}{2}(c_5-c_6) +\cdots \\ = \color{red}{\dfrac{5}{2\times 3} - \dfrac{11\cdot 1!}{2\times 3 \cdot 5 \cdot 7} + \dfrac{17\cdot 2!}{2\times 5 \cdot 7 \cdot 9 \cdot 11} } - \dfrac{3!}{7 \cdot 9 \cdot 11 \cdot 13} + \dfrac{3!}{9 \cdot 11 \cdot 13 \cdot 15} - \cdots $
$~$
See this paper for more info.
This transformation can be described by formula:
for alternate series
$$ S = \sum_{n=0}^{\infty} u_n, $$ improved series has form $$ S = \sum_{n=0}^{\infty} s_n = \sum_{n=0}^{\infty} \left(\dfrac{\Delta^n u_n}{2^n} + \dfrac{\Delta^n u_{n+1}}{2^{n+1}}\right), $$
where $$ \Delta^n u_j = \sum_{k=0}^n \binom{n}{k} u_{j+k}. $$
Really, if we have $u_0=1,u_1=-1/3, u_2=1/5, u_3=-1/7, ...$, then
$s_0 = \dfrac{u_0}{1} + \dfrac{u_1}{2} = 1-\dfrac{1}{2\cdot 3} = \dfrac{5}{6}$,
$s_1 = \dfrac{u_1+u_2}{2} + \dfrac{u_2+u_3}{4} = \dfrac{-\frac{1}{3}+\frac{1}{5}}{2} + \dfrac{\frac{1}{5}-\frac{1}{7}}{4} = -\dfrac{11}{210}$,
$s_2 = \dfrac{u_2+2u_3+u_4}{4} + \dfrac{u_3+2u_4+u_5}{8} = \dfrac{\frac{1}{5}-\frac{2}{7}+\frac{1}{9}}{4} + \dfrac{-\frac{1}{7}+\frac{2}{9}-\frac{1}{11}}{8} = \dfrac{17}{3465}$;
...
For completeness, I'll show "table of weights" for Euler's transform, and for this modified transform.
$$ s_j = \sum_k w_{jk} a_k. $$
Euler's transform:
\begin{array}{|c|c|c|c|c|c|c|c|} \hline & a_0 & a_1 & a_2 & a_3 & a_4 & a_5 & \cdots \\ \hline s_0: & \dfrac{1}{2} & & & & & & \\ \hline s_1: & \dfrac{1}{4} & \dfrac{1}{4} & & & & & \\ \hline s_2: & \dfrac{1}{8} & \dfrac{2}{8} & \dfrac{1}{8} & & & & \\ \hline s_3: & \dfrac{1}{16} & \dfrac{3}{16} & \dfrac{3}{16} & \dfrac{1}{16} & & & \\ \hline s_4: & \dfrac{1}{32} & \dfrac{4}{32} & \dfrac{6}{32} & \dfrac{4}{32} & \dfrac{1}{32} & & \\ \hline s_5: & \dfrac{1}{64} & \dfrac{5}{64} & \dfrac{10}{64} & \dfrac{10}{64} & \dfrac{5}{64} & \dfrac{1}{64} & \\ \hline \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots \\ \hline \\ \sum_j w_{jk} & 1 & 1 & 1 & 1 & 1 & 1 & ... \\ \hline \end{array}
This transform:
$ \begin{array}{|c|c|c|c|c|c|c|c|c|c|c|} \hline & a_0 & a_1 & a_2 & a_3 & a_4 & a_5 & a_6 & a_7 & a_8 & \cdots \\ \hline s_0: & 1 & \frac{1}{2} \\ \hline s_1: & & \frac{1}{2} & \frac{1}{2}+ \frac{1}{4} & \frac{1}{4} \\ \hline s_2: & & & \frac{1}{4} & \frac{2}{4} + \frac{1}{8} & \frac{1}{4}+\frac{2}{8} & \frac{1}{8} \\ \hline s_3: & & & & \frac{1}{8}& \frac{3}{8} + \frac{1}{16} & \frac{3}{8}+\frac{3}{16} & \frac{1}{8} +\frac{3}{16} & \frac{1}{16} \\ \hline s_4: & & & & & \frac{1}{16} & \frac{4}{16} + \frac{1}{32} & \frac{6}{16}+\frac{4}{32} & \frac{4}{16} +\frac{6}{32} & \frac{1}{16}+\frac{4}{32} & \frac{1}{32} \\ \hline \vdots & & & & & & \ddots & \vdots & \vdots& \vdots & \vdots \\ \hline \\ \displaystyle\sum\limits_j w_{jk} & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & \cdots \\ \hline \end{array} $