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
asked Nov 19 at 19:12
Joey Marlowe
966
add a comment |
up vote
4
down vote
favorite
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
asked Nov 19 at 19:12
Joey Marlowe
966
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
Nov 19 at 20:05
add a comment |
up vote
4
down vote
favorite
up vote
4
down vote
favorite
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
asked Nov 19 at 19:12
Joey Marlowe
966
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
asked Nov 19 at 19:12
Joey Marlowe
966
asked Nov 19 at 19:12
Joey Marlowe
966
edited Nov 19 at 19:45
asked Nov 19 at 19:12
Joey Marlowe
966
asked Nov 19 at 19:12
Joey Marlowe
966
asked Nov 19 at 19:12
Joey Marlowe
966
966
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
Nov 19 at 20:05
add a comment |
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
Nov 19 at 20:05
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
Nov 19 at 20:05
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
Nov 19 at 20:05
add a comment |
1 Answer
1
active
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votes
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
$$S(m)/pi={1 over 2}, ; { -1 over 4}, ; { 1 over 16}, ; { -1 over 96}, ; { 1 over 768}, ; { -1 over 7680}, ; { 1 over 92160},, cdots$$
answered Nov 20 at 1:41
G Cab
17.3k31237
Wow. Thank you!
– Joey Marlowe
Nov 20 at 2:24
1
@JoeyMarlowe: glad to have helped and solve the ..mistery
– G Cab
Nov 20 at 14:52
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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
$$S(m)/pi={1 over 2}, ; { -1 over 4}, ; { 1 over 16}, ; { -1 over 96}, ; { 1 over 768}, ; { -1 over 7680}, ; { 1 over 92160},, cdots$$
answered Nov 20 at 1:41
G Cab
17.3k31237
Wow. Thank you!
– Joey Marlowe
Nov 20 at 2:24
1
@JoeyMarlowe: glad to have helped and solve the ..mistery
– G Cab
Nov 20 at 14:52
add a comment |
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
$$S(m)/pi={1 over 2}, ; { -1 over 4}, ; { 1 over 16}, ; { -1 over 96}, ; { 1 over 768}, ; { -1 over 7680}, ; { 1 over 92160},, cdots$$
answered Nov 20 at 1:41
G Cab
17.3k31237
Wow. Thank you!
– Joey Marlowe
Nov 20 at 2:24
1
@JoeyMarlowe: glad to have helped and solve the ..mistery
– G Cab
Nov 20 at 14:52
add a comment |
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
$$S(m)/pi={1 over 2}, ; { -1 over 4}, ; { 1 over 16}, ; { -1 over 96}, ; { 1 over 768}, ; { -1 over 7680}, ; { 1 over 92160},, cdots$$
answered Nov 20 at 1:41
G Cab
17.3k31237
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
$$S(m)/pi={1 over 2}, ; { -1 over 4}, ; { 1 over 16}, ; { -1 over 96}, ; { 1 over 768}, ; { -1 over 7680}, ; { 1 over 92160},, cdots$$
answered Nov 20 at 1:41
G Cab
17.3k31237
edited Nov 20 at 23:26
answered Nov 20 at 1:41
G Cab
17.3k31237
answered Nov 20 at 1:41
G Cab
17.3k31237
answered Nov 20 at 1:41
G Cab
17.3k31237
17.3k31237
Wow. Thank you!
– Joey Marlowe
Nov 20 at 2:24
1
@JoeyMarlowe: glad to have helped and solve the ..mistery
– G Cab
Nov 20 at 14:52
add a comment |
Wow. Thank you!
– Joey Marlowe
Nov 20 at 2:24
1
@JoeyMarlowe: glad to have helped and solve the ..mistery
– G Cab
Nov 20 at 14:52
Wow. Thank you!
– Joey Marlowe
Nov 20 at 2:24
Wow. Thank you!
– Joey Marlowe
Nov 20 at 2:24
1
1
@JoeyMarlowe: glad to have helped and solve the ..mistery
– G Cab
Nov 20 at 14:52
@JoeyMarlowe: glad to have helped and solve the ..mistery
– G Cab
Nov 20 at 14:52
add a comment |
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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
Nov 19 at 20:05