TITLE: Help me ID this weird $\pi$ formula
QUESTION [8 upvotes]: 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?

REPLY [6 votes]: 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}
$