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











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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:
$$ frac{pi}{2} = sum_{n=1}^infty frac{(2n-5)!!}{(2n-3)(2n-2)!!}$$






sequences-and-series convergence taylor-expansion factorial pi







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  • 2




    It appears that the general formula is $sum_{n=0}^{infty}frac{(2(n-k)-1)!!}{(2n-1)(2n)!!}=frac{(-1)^kpi}{2(2k)!!}$
    – Jacob
    2 days ago















up vote
4
down vote

favorite
2












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:
$$ frac{pi}{2} = sum_{n=1}^infty frac{(2n-5)!!}{(2n-3)(2n-2)!!}$$






sequences-and-series convergence taylor-expansion factorial pi







share|cite|improve this question




















  • 2




    It appears that the general formula is $sum_{n=0}^{infty}frac{(2(n-k)-1)!!}{(2n-1)(2n)!!}=frac{(-1)^kpi}{2(2k)!!}$
    – Jacob
    2 days ago













up vote
4
down vote

favorite
2









up vote
4
down vote

favorite
2






2





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:
$$ frac{pi}{2} = sum_{n=1}^infty frac{(2n-5)!!}{(2n-3)(2n-2)!!}$$






sequences-and-series convergence taylor-expansion factorial pi







share|cite|improve this 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:
$$ frac{pi}{2} = sum_{n=1}^infty frac{(2n-5)!!}{(2n-3)(2n-2)!!}$$






sequences-and-series convergence taylor-expansion factorial pi




sequences-and-series convergence taylor-expansion factorial pi






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share|cite|improve this question













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edited 2 days ago

























asked 2 days ago









Joey Marlowe

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  • 2




    It appears that the general formula is $sum_{n=0}^{infty}frac{(2(n-k)-1)!!}{(2n-1)(2n)!!}=frac{(-1)^kpi}{2(2k)!!}$
    – Jacob
    2 days ago














  • 2




    It appears that the general formula is $sum_{n=0}^{infty}frac{(2(n-k)-1)!!}{(2n-1)(2n)!!}=frac{(-1)^kpi}{2(2k)!!}$
    – Jacob
    2 days ago








2




2




It appears that the general formula is $sum_{n=0}^{infty}frac{(2(n-k)-1)!!}{(2n-1)(2n)!!}=frac{(-1)^kpi}{2(2k)!!}$
– Jacob
2 days ago




It appears that the general formula is $sum_{n=0}^{infty}frac{(2(n-k)-1)!!}{(2n-1)(2n)!!}=frac{(-1)^kpi}{2(2k)!!}$
– Jacob
2 days ago










1 Answer
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up vote
2
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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
$$
left( {2n - 3} right)!! = left( {2left( {n - 2} right) + 1} right)!! = 2^{,n - 1} left( {{1 over 2}} right)^{,overline {,n - 1,} }
= {{1^{,overline {,2left( {n - 1} right),} } } over {2^{,n - 1} 1^{,overline {,n - 1,} } }}
$$



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
$$
Gamma left( z right),Gamma left( { - z} right) = - {pi over {zsin left( {pi ,z} right)}}
$$
- 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{
& sumlimits_{1, le ,n} {{{left( {left( {2n - 3} right)!!} right)^{,2} } over {left( {2n - 3} right)left( {2n - 1} right)!}}}
= sumlimits_{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
& = sumlimits_{0, le ,n} {{{1^{,overline {,2n,} } } over {2^{,2n} ,1^{,overline {,n,} } ,1^{,overline {,n,} } left( {2n - 1} right)
left( {2n + 1} right)}}} = cr
& = sumlimits_{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
& = sumlimits_{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 }}sumlimits_{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 {4sqrt pi }}sumlimits_{0, le ,n} {{{Gamma left( {n - 1/2} right)} over {Gamma left( {n + 1} right)
left( {n + 1/2} right)}}} = cr
& = {1 over {4sqrt pi }}sumlimits_{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 {4sqrt pi }}{{Gamma left( {1/2} right)Gamma left( { - 1/2} right)} over {Gamma left( {3/2} right)}}
sumlimits_{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
& = - {{2pi } over {4sqrt pi }}{2 over {sqrt pi }}sumlimits_{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
$$ bbox[lightyellow] {
eqalign{
& S(m) = sumlimits_{1, le ,n} {{{left( {2n - 2m - 1} right)!!} over {left( {2n - 1} right)left( {2n - 2} right)!!}}}
= sumlimits_{0, le ,n} {{{left( {2n - 2m + 1} right)!!} over {left( {2n + 1} right)left( {2n} right)!!}}} = cr
& = {{2^{, - m} } over {sqrt pi }}sumlimits_{0, le ,n} {{{Gamma left( {n - m + 3/2} right)} over {left( {n + 1/2} right)}}{1 over {n!}}} = cr
& = {{2^{, - m + 1} Gamma left( { - m + 3/2} right)} over {sqrt pi }}sumlimits_{0, le ,n}
{{{left( {1/2} right)^{,overline {,n,} } left( { - m + 3/2} right)^{,overline {,n,} } } over {left( {3/2} right)^{,overline {,n,} } }}{1 over {n!}}} = cr
& = {{2^{, - m + 1} Gamma left( { - m + 3/2} right)} over {sqrt pi }} ,
{}_2F_{,1} left( {left. {matrix{ {1/2,3/2 - m} cr {3/2} cr } ,} right|;1} right) = cr
& = {{2^{, - m + 1} Gamma left( { - m + 3/2} right)} over {sqrt pi }}{{Gamma left( {3/2} right)Gamma left( {m - 1/2} right)}
over {Gamma left( 1 right)Gamma left( m right)}} = cr
& = - {1 over {;2^{,m} Gamma left( m right)cos left( {pi m} right)}};pi quad left| {;1/2 < {mathop{rm Re}nolimits} (m)} right. cr}
}$$
valid also for complex $m$.



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






share|cite|improve this answer























  • Wow. Thank you!
    – Joey Marlowe
    yesterday






  • 1




    @JoeyMarlowe: glad to have helped and solve the ..mistery
    – G Cab
    yesterday











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1 Answer
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1 Answer
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active

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up vote
2
down vote



accepted










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
$$
left( {2n - 3} right)!! = left( {2left( {n - 2} right) + 1} right)!! = 2^{,n - 1} left( {{1 over 2}} right)^{,overline {,n - 1,} }
= {{1^{,overline {,2left( {n - 1} right),} } } over {2^{,n - 1} 1^{,overline {,n - 1,} } }}
$$



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
$$
Gamma left( z right),Gamma left( { - z} right) = - {pi over {zsin left( {pi ,z} right)}}
$$
- 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{
& sumlimits_{1, le ,n} {{{left( {left( {2n - 3} right)!!} right)^{,2} } over {left( {2n - 3} right)left( {2n - 1} right)!}}}
= sumlimits_{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
& = sumlimits_{0, le ,n} {{{1^{,overline {,2n,} } } over {2^{,2n} ,1^{,overline {,n,} } ,1^{,overline {,n,} } left( {2n - 1} right)
left( {2n + 1} right)}}} = cr
& = sumlimits_{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
& = sumlimits_{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 }}sumlimits_{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 {4sqrt pi }}sumlimits_{0, le ,n} {{{Gamma left( {n - 1/2} right)} over {Gamma left( {n + 1} right)
left( {n + 1/2} right)}}} = cr
& = {1 over {4sqrt pi }}sumlimits_{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 {4sqrt pi }}{{Gamma left( {1/2} right)Gamma left( { - 1/2} right)} over {Gamma left( {3/2} right)}}
sumlimits_{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
& = - {{2pi } over {4sqrt pi }}{2 over {sqrt pi }}sumlimits_{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
$$ bbox[lightyellow] {
eqalign{
& S(m) = sumlimits_{1, le ,n} {{{left( {2n - 2m - 1} right)!!} over {left( {2n - 1} right)left( {2n - 2} right)!!}}}
= sumlimits_{0, le ,n} {{{left( {2n - 2m + 1} right)!!} over {left( {2n + 1} right)left( {2n} right)!!}}} = cr
& = {{2^{, - m} } over {sqrt pi }}sumlimits_{0, le ,n} {{{Gamma left( {n - m + 3/2} right)} over {left( {n + 1/2} right)}}{1 over {n!}}} = cr
& = {{2^{, - m + 1} Gamma left( { - m + 3/2} right)} over {sqrt pi }}sumlimits_{0, le ,n}
{{{left( {1/2} right)^{,overline {,n,} } left( { - m + 3/2} right)^{,overline {,n,} } } over {left( {3/2} right)^{,overline {,n,} } }}{1 over {n!}}} = cr
& = {{2^{, - m + 1} Gamma left( { - m + 3/2} right)} over {sqrt pi }} ,
{}_2F_{,1} left( {left. {matrix{ {1/2,3/2 - m} cr {3/2} cr } ,} right|;1} right) = cr
& = {{2^{, - m + 1} Gamma left( { - m + 3/2} right)} over {sqrt pi }}{{Gamma left( {3/2} right)Gamma left( {m - 1/2} right)}
over {Gamma left( 1 right)Gamma left( m right)}} = cr
& = - {1 over {;2^{,m} Gamma left( m right)cos left( {pi m} right)}};pi quad left| {;1/2 < {mathop{rm Re}nolimits} (m)} right. cr}
}$$
valid also for complex $m$.



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






share|cite|improve this answer























  • Wow. Thank you!
    – Joey Marlowe
    yesterday






  • 1




    @JoeyMarlowe: glad to have helped and solve the ..mistery
    – G Cab
    yesterday















up vote
2
down vote



accepted










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
$$
left( {2n - 3} right)!! = left( {2left( {n - 2} right) + 1} right)!! = 2^{,n - 1} left( {{1 over 2}} right)^{,overline {,n - 1,} }
= {{1^{,overline {,2left( {n - 1} right),} } } over {2^{,n - 1} 1^{,overline {,n - 1,} } }}
$$



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
$$
Gamma left( z right),Gamma left( { - z} right) = - {pi over {zsin left( {pi ,z} right)}}
$$
- 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{
& sumlimits_{1, le ,n} {{{left( {left( {2n - 3} right)!!} right)^{,2} } over {left( {2n - 3} right)left( {2n - 1} right)!}}}
= sumlimits_{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
& = sumlimits_{0, le ,n} {{{1^{,overline {,2n,} } } over {2^{,2n} ,1^{,overline {,n,} } ,1^{,overline {,n,} } left( {2n - 1} right)
left( {2n + 1} right)}}} = cr
& = sumlimits_{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
& = sumlimits_{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 }}sumlimits_{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 {4sqrt pi }}sumlimits_{0, le ,n} {{{Gamma left( {n - 1/2} right)} over {Gamma left( {n + 1} right)
left( {n + 1/2} right)}}} = cr
& = {1 over {4sqrt pi }}sumlimits_{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 {4sqrt pi }}{{Gamma left( {1/2} right)Gamma left( { - 1/2} right)} over {Gamma left( {3/2} right)}}
sumlimits_{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
& = - {{2pi } over {4sqrt pi }}{2 over {sqrt pi }}sumlimits_{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
$$ bbox[lightyellow] {
eqalign{
& S(m) = sumlimits_{1, le ,n} {{{left( {2n - 2m - 1} right)!!} over {left( {2n - 1} right)left( {2n - 2} right)!!}}}
= sumlimits_{0, le ,n} {{{left( {2n - 2m + 1} right)!!} over {left( {2n + 1} right)left( {2n} right)!!}}} = cr
& = {{2^{, - m} } over {sqrt pi }}sumlimits_{0, le ,n} {{{Gamma left( {n - m + 3/2} right)} over {left( {n + 1/2} right)}}{1 over {n!}}} = cr
& = {{2^{, - m + 1} Gamma left( { - m + 3/2} right)} over {sqrt pi }}sumlimits_{0, le ,n}
{{{left( {1/2} right)^{,overline {,n,} } left( { - m + 3/2} right)^{,overline {,n,} } } over {left( {3/2} right)^{,overline {,n,} } }}{1 over {n!}}} = cr
& = {{2^{, - m + 1} Gamma left( { - m + 3/2} right)} over {sqrt pi }} ,
{}_2F_{,1} left( {left. {matrix{ {1/2,3/2 - m} cr {3/2} cr } ,} right|;1} right) = cr
& = {{2^{, - m + 1} Gamma left( { - m + 3/2} right)} over {sqrt pi }}{{Gamma left( {3/2} right)Gamma left( {m - 1/2} right)}
over {Gamma left( 1 right)Gamma left( m right)}} = cr
& = - {1 over {;2^{,m} Gamma left( m right)cos left( {pi m} right)}};pi quad left| {;1/2 < {mathop{rm Re}nolimits} (m)} right. cr}
}$$
valid also for complex $m$.



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






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  • Wow. Thank you!
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up vote
2
down vote



accepted







up vote
2
down vote



accepted






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
$$
left( {2n - 3} right)!! = left( {2left( {n - 2} right) + 1} right)!! = 2^{,n - 1} left( {{1 over 2}} right)^{,overline {,n - 1,} }
= {{1^{,overline {,2left( {n - 1} right),} } } over {2^{,n - 1} 1^{,overline {,n - 1,} } }}
$$



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
$$
Gamma left( z right),Gamma left( { - z} right) = - {pi over {zsin left( {pi ,z} right)}}
$$
- 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{
& sumlimits_{1, le ,n} {{{left( {left( {2n - 3} right)!!} right)^{,2} } over {left( {2n - 3} right)left( {2n - 1} right)!}}}
= sumlimits_{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
& = sumlimits_{0, le ,n} {{{1^{,overline {,2n,} } } over {2^{,2n} ,1^{,overline {,n,} } ,1^{,overline {,n,} } left( {2n - 1} right)
left( {2n + 1} right)}}} = cr
& = sumlimits_{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
& = sumlimits_{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 }}sumlimits_{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 {4sqrt pi }}sumlimits_{0, le ,n} {{{Gamma left( {n - 1/2} right)} over {Gamma left( {n + 1} right)
left( {n + 1/2} right)}}} = cr
& = {1 over {4sqrt pi }}sumlimits_{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 {4sqrt pi }}{{Gamma left( {1/2} right)Gamma left( { - 1/2} right)} over {Gamma left( {3/2} right)}}
sumlimits_{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
& = - {{2pi } over {4sqrt pi }}{2 over {sqrt pi }}sumlimits_{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
$$ bbox[lightyellow] {
eqalign{
& S(m) = sumlimits_{1, le ,n} {{{left( {2n - 2m - 1} right)!!} over {left( {2n - 1} right)left( {2n - 2} right)!!}}}
= sumlimits_{0, le ,n} {{{left( {2n - 2m + 1} right)!!} over {left( {2n + 1} right)left( {2n} right)!!}}} = cr
& = {{2^{, - m} } over {sqrt pi }}sumlimits_{0, le ,n} {{{Gamma left( {n - m + 3/2} right)} over {left( {n + 1/2} right)}}{1 over {n!}}} = cr
& = {{2^{, - m + 1} Gamma left( { - m + 3/2} right)} over {sqrt pi }}sumlimits_{0, le ,n}
{{{left( {1/2} right)^{,overline {,n,} } left( { - m + 3/2} right)^{,overline {,n,} } } over {left( {3/2} right)^{,overline {,n,} } }}{1 over {n!}}} = cr
& = {{2^{, - m + 1} Gamma left( { - m + 3/2} right)} over {sqrt pi }} ,
{}_2F_{,1} left( {left. {matrix{ {1/2,3/2 - m} cr {3/2} cr } ,} right|;1} right) = cr
& = {{2^{, - m + 1} Gamma left( { - m + 3/2} right)} over {sqrt pi }}{{Gamma left( {3/2} right)Gamma left( {m - 1/2} right)}
over {Gamma left( 1 right)Gamma left( m right)}} = cr
& = - {1 over {;2^{,m} Gamma left( m right)cos left( {pi m} right)}};pi quad left| {;1/2 < {mathop{rm Re}nolimits} (m)} right. cr}
}$$
valid also for complex $m$.



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






share|cite|improve this answer














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
$$
left( {2n - 3} right)!! = left( {2left( {n - 2} right) + 1} right)!! = 2^{,n - 1} left( {{1 over 2}} right)^{,overline {,n - 1,} }
= {{1^{,overline {,2left( {n - 1} right),} } } over {2^{,n - 1} 1^{,overline {,n - 1,} } }}
$$



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
$$
Gamma left( z right),Gamma left( { - z} right) = - {pi over {zsin left( {pi ,z} right)}}
$$
- 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{
& sumlimits_{1, le ,n} {{{left( {left( {2n - 3} right)!!} right)^{,2} } over {left( {2n - 3} right)left( {2n - 1} right)!}}}
= sumlimits_{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
& = sumlimits_{0, le ,n} {{{1^{,overline {,2n,} } } over {2^{,2n} ,1^{,overline {,n,} } ,1^{,overline {,n,} } left( {2n - 1} right)
left( {2n + 1} right)}}} = cr
& = sumlimits_{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
& = sumlimits_{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 }}sumlimits_{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 {4sqrt pi }}sumlimits_{0, le ,n} {{{Gamma left( {n - 1/2} right)} over {Gamma left( {n + 1} right)
left( {n + 1/2} right)}}} = cr
& = {1 over {4sqrt pi }}sumlimits_{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 {4sqrt pi }}{{Gamma left( {1/2} right)Gamma left( { - 1/2} right)} over {Gamma left( {3/2} right)}}
sumlimits_{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
& = - {{2pi } over {4sqrt pi }}{2 over {sqrt pi }}sumlimits_{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
$$ bbox[lightyellow] {
eqalign{
& S(m) = sumlimits_{1, le ,n} {{{left( {2n - 2m - 1} right)!!} over {left( {2n - 1} right)left( {2n - 2} right)!!}}}
= sumlimits_{0, le ,n} {{{left( {2n - 2m + 1} right)!!} over {left( {2n + 1} right)left( {2n} right)!!}}} = cr
& = {{2^{, - m} } over {sqrt pi }}sumlimits_{0, le ,n} {{{Gamma left( {n - m + 3/2} right)} over {left( {n + 1/2} right)}}{1 over {n!}}} = cr
& = {{2^{, - m + 1} Gamma left( { - m + 3/2} right)} over {sqrt pi }}sumlimits_{0, le ,n}
{{{left( {1/2} right)^{,overline {,n,} } left( { - m + 3/2} right)^{,overline {,n,} } } over {left( {3/2} right)^{,overline {,n,} } }}{1 over {n!}}} = cr
& = {{2^{, - m + 1} Gamma left( { - m + 3/2} right)} over {sqrt pi }} ,
{}_2F_{,1} left( {left. {matrix{ {1/2,3/2 - m} cr {3/2} cr } ,} right|;1} right) = cr
& = {{2^{, - m + 1} Gamma left( { - m + 3/2} right)} over {sqrt pi }}{{Gamma left( {3/2} right)Gamma left( {m - 1/2} right)}
over {Gamma left( 1 right)Gamma left( m right)}} = cr
& = - {1 over {;2^{,m} Gamma left( m right)cos left( {pi m} right)}};pi quad left| {;1/2 < {mathop{rm Re}nolimits} (m)} right. cr}
}$$
valid also for complex $m$.



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







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited yesterday

























answered yesterday









G Cab

16.9k31237




16.9k31237












  • Wow. Thank you!
    – Joey Marlowe
    yesterday






  • 1




    @JoeyMarlowe: glad to have helped and solve the ..mistery
    – G Cab
    yesterday


















  • Wow. Thank you!
    – Joey Marlowe
    yesterday






  • 1




    @JoeyMarlowe: glad to have helped and solve the ..mistery
    – G Cab
    yesterday
















Wow. Thank you!
– Joey Marlowe
yesterday




Wow. Thank you!
– Joey Marlowe
yesterday




1




1




@JoeyMarlowe: glad to have helped and solve the ..mistery
– G Cab
yesterday




@JoeyMarlowe: glad to have helped and solve the ..mistery
– G Cab
yesterday


















 

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