Help understanding the cause of this pattern when writing π as an infinite series with double factorials

Question

I made a post about a year and a half ago: $\pi$ as an Infinite Series using Taylor Expansion on Equation of a Circle where essentially I used the Taylor series expansion on $\ y = \sqrt{r^2-x^2}$ (the equation of a circle in Cartesian coordinates with radius $r$). Integrating term by term from $0$ to $r$ in order to obtain a representation of $\pi$ gave a pattern which I wrote as: $$\ \pi = \sum_{n=1}^\infty \frac{-4[(2n-3)!!]^2}{(2n-3)(2n-1)!}$$ This can be written differently as: $$\ -\frac{\pi}{4} = \sum_{n=1}^\infty \frac{(2n-5)!!}{(2n-1)(2n-2)!!}$$ Furthermore what I have noticed is (and what I'm trying to understand but cannot for some reason): $$\ \frac{\pi}{16} = \sum_{n=1}^\infty \frac{(2n-7)!!}{(2n-1)(2n-2)!!}$$ $$\ -\frac{\pi}{96} = \sum_{n=1}^\infty \frac{(2n-9)!!}{(2n-1)(2n-2)!!}$$ $$\ \frac{\pi}{768} = \sum_{n=1}^\infty \frac{(2n-11)!!}{(2n-1)(2n-2)!!}$$ This pattern continues when adjusting the numerator (double factorial term) with odd integers... 5, 7, 9, 11, 13, and so on. Those series will converge to fractions of pi and alternate as positive or negative fractions of pi. I do not understand what is happening here. Also, it is interesting to note that it seems: $$\ \frac{\pi}{2} = \sum_{n=1}^\infty \frac{(2n-5)!!}{(2n-3)(2n-2)!!}$$

Answer 1

Considering the definition of the Rising Factorial $$ z^{\,\overline {\,n\,} } = {{\Gamma \left( {z + n} \right)} \over {\Gamma \left( z \right)}} $$ the double factorial can then be written as $$ \left( {2n - 3} \right)!! = \left( {2\left( {n - 2} \right) + 1} \right)!! = 2^{\,n - 1} \left( {{1 \over 2}} \right)^{\,\overline {\,n - 1\,} } = {{1^{\,\overline {\,2\left( {n - 1} \right)\,} } } \over {2^{\,n - 1} 1^{\,\overline {\,n - 1\,} } }} $$

We apply :
- the Gamma Duplication formula $$ \Gamma \left( {2\,z} \right) = {{2^{\,2\,z - 1} } \over {\sqrt \pi }}\Gamma \left( z \right)\Gamma \left( {z + 1/2} \right)\quad $$ - the Gamma Reflection formula $$ \Gamma \left( z \right)\,\Gamma \left( { - z} \right) = - {\pi \over {z\sin \left( {\pi \,z} \right)}} $$ - and the Gauss theorem for the Hypergeometric function $$ {}_2F_{\,1} \left( {\left. {\matrix{ {a,b} \cr c \cr } \,} \right|\;1} \right) = {{\Gamma \left( c \right)\Gamma \left( {c - a - b} \right)} \over {\Gamma \left( {c - a} \right)\Gamma \left( {c - b} \right)}} \quad \left| {\;{\mathop{\rm Re}\nolimits} (a + b) < {\mathop{\rm Re}\nolimits} (c)} \right. $$

Then the sum becomes $$ \eqalign{ & \sum\limits_{1\, \le \,n} {{{\left( {\left( {2n - 3} \right)!!} \right)^{\,2} } \over {\left( {2n - 3} \right)\left( {2n - 1} \right)!}}} = \sum\limits_{0\, \le \,n} {{{1^{\,\overline {\,2n\,} } 1^{\,\overline {\,2n\,} } } \over {2^{\,2n} \,1^{\,\overline {\,n\,} } \,1^{\,\overline {\,n\,} } \left( {2n - 1} \right)1^{\,\overline {\,2n + 1\,} } }}} = \cr & = \sum\limits_{0\, \le \,n} {{{1^{\,\overline {\,2n\,} } } \over {2^{\,2n} \,1^{\,\overline {\,n\,} } \,1^{\,\overline {\,n\,} } \left( {2n - 1} \right) \left( {2n + 1} \right)}}} = \cr & = \sum\limits_{0\, \le \,n} {{{\Gamma \left( {2n + 1} \right)} \over {2^{\,2n} \Gamma \left( {n + 1} \right)\Gamma \left( {n + 1} \right) \left( {2n - 1} \right)\left( {2n + 1} \right)}}} = \cr & = \sum\limits_{0\, \le \,n} {{{2n{{2^{\,2\,n - 1} } \over {\sqrt \pi }}\Gamma \left( n \right)\Gamma \left( {n + 1/2} \right)} \over {2^{\,2n} \Gamma \left( {n + 1} \right)\Gamma \left( {n + 1} \right)\left( {2n - 1} \right)\left( {2n + 1} \right)}}} = \cr & = {1 \over {\sqrt \pi }}\sum\limits_{0\, \le \,n} {{{\Gamma \left( {n + 1/2} \right)} \over {\Gamma \left( {n + 1} \right)\left( {2n - 1} \right) \left( {2n + 1} \right)}}} = \cr & = {1 \over {4\sqrt \pi }}\sum\limits_{0\, \le \,n} {{{\Gamma \left( {n - 1/2} \right)} \over {\Gamma \left( {n + 1} \right) \left( {n + 1/2} \right)}}} = \cr & = {1 \over {4\sqrt \pi }}\sum\limits_{0\, \le \,n} {{{\Gamma \left( {n + 1/2} \right)\Gamma \left( {n - 1/2} \right)} \over {\Gamma \left( {n + 3/2} \right)}}{1 \over {n!}}} = \cr & = {1 \over {4\sqrt \pi }}{{\Gamma \left( {1/2} \right)\Gamma \left( { - 1/2} \right)} \over {\Gamma \left( {3/2} \right)}} \sum\limits_{0\, \le \,n} {{{\left( {1/2} \right)^{\,\overline {\,n\,} } \left( { - 1/2} \right)^{\,\overline {\,n\,} } } \over {\left( {3/2} \right)^{\,\overline {\,n\,} } }}{1 \over {n!}}} = \cr & = - {{2\pi } \over {4\sqrt \pi }}{2 \over {\sqrt \pi }}\sum\limits_{0\, \le \,n} {{{\left( {1/2} \right)^{\,\overline {\,n\,} } \left( { - 1/2} \right)^{\,\overline {\,n\,} } } \over {\left( {3/2} \right)^{\,\overline {\,n\,} } }}{1 \over {n!}}} = \cr & = - {}_2F_{\,1} \left( {\left. {\matrix{ {1/2, - 1/2} \cr {3/2} \cr } \,} \right|\;1} \right) = - {{\Gamma \left( {3/2} \right)\Gamma \left( {3/2} \right)} \over {\Gamma \left( 1 \right)\Gamma \left( 2 \right)}} = \cr & = - {\pi \over 4} \cr} $$

The other sums follow a similar pattern.

In fact, following the same steps as above (here reported concisely) we reach to the general expression $$ \bbox[lightyellow] { \eqalign{ & S(m) = \sum\limits_{1\, \le \,n} {{{\left( {2n - 2m - 1} \right)!!} \over {\left( {2n - 1} \right)\left( {2n - 2} \right)!!}}} = \sum\limits_{0\, \le \,n} {{{\left( {2n - 2m + 1} \right)!!} \over {\left( {2n + 1} \right)\left( {2n} \right)!!}}} = \cr & = {{2^{\, - m} } \over {\sqrt \pi }}\sum\limits_{0\, \le \,n} {{{\Gamma \left( {n - m + 3/2} \right)} \over {\left( {n + 1/2} \right)}}{1 \over {n!}}} = \cr & = {{2^{\, - m + 1} \Gamma \left( { - m + 3/2} \right)} \over {\sqrt \pi }}\sum\limits_{0\, \le \,n} {{{\left( {1/2} \right)^{\,\overline {\,n\,} } \left( { - m + 3/2} \right)^{\,\overline {\,n\,} } } \over {\left( {3/2} \right)^{\,\overline {\,n\,} } }}{1 \over {n!}}} = \cr & = {{2^{\, - m + 1} \Gamma \left( { - m + 3/2} \right)} \over {\sqrt \pi }} \, {}_2F_{\,1} \left( {\left. {\matrix{ {1/2,3/2 - m} \cr {3/2} \cr } \,} \right|\;1} \right) = \cr & = {{2^{\, - m + 1} \Gamma \left( { - m + 3/2} \right)} \over {\sqrt \pi }}{{\Gamma \left( {3/2} \right)\Gamma \left( {m - 1/2} \right)} \over {\Gamma \left( 1 \right)\Gamma \left( m \right)}} = \cr & = - {1 \over {\;2^{\,m} \Gamma \left( m \right)\cos \left( {\pi m} \right)}}\;\pi \quad \left| {\;1/2 < {\mathop{\rm Re}\nolimits} (m)} \right. \cr} }$$ valid also for complex $m$.

For integer $m=1,2, \cdots, 7$ that gives $$S(m)/\pi={1 \over 2}, \; { -1 \over 4}, \; { 1 \over 16}, \; { -1 \over 96}, \; { 1 \over 768}, \; { -1 \over 7680}, \; { 1 \over 92160},\, \cdots$$