Mixing
Verifying $ sum_{k=0}^{infty}{3k choose k}frac{9k^2-3k-1}{(3k-1)(3k-2)}left(frac{2}{27}right)^k=frac{1}{4}$
$begingroup$
$$
sum_{k=0}^{infty}{3k choose k}dfrac{9k^2-3k-1}{(3k-1)(3k-2)}left(dfrac{2}{27}right)^k=dfrac{1}{4}
$$
After some simplification, I got the following result:
$$
sum_{k=0}^{infty}left{{3k choose k}+dfrac{27}{2k}{3k-4 choose 2k-3}right}left(dfrac{2}{27}right)^k.
$$
Now, how I can proceed?
real-analysis summation
edited Jan 20 at 18:27
Blue
48.7k870156
asked Jan 20 at 11:52
J.DoeJ.Doe
311
$endgroup$
add a comment |
$begingroup$
$$
sum_{k=0}^{infty}{3k choose k}dfrac{9k^2-3k-1}{(3k-1)(3k-2)}left(dfrac{2}{27}right)^k=dfrac{1}{4}
$$
After some simplification, I got the following result:
$$
sum_{k=0}^{infty}left{{3k choose k}+dfrac{27}{2k}{3k-4 choose 2k-3}right}left(dfrac{2}{27}right)^k.
$$
Now, how I can proceed?
real-analysis summation
edited Jan 20 at 18:27
Blue
48.7k870156
asked Jan 20 at 11:52
J.DoeJ.Doe
311
$endgroup$
add a comment |
$begingroup$
$$
sum_{k=0}^{infty}{3k choose k}dfrac{9k^2-3k-1}{(3k-1)(3k-2)}left(dfrac{2}{27}right)^k=dfrac{1}{4}
$$
After some simplification, I got the following result:
$$
sum_{k=0}^{infty}left{{3k choose k}+dfrac{27}{2k}{3k-4 choose 2k-3}right}left(dfrac{2}{27}right)^k.
$$
Now, how I can proceed?
real-analysis summation
edited Jan 20 at 18:27
Blue
48.7k870156
asked Jan 20 at 11:52
J.DoeJ.Doe
311
$endgroup$
$$
sum_{k=0}^{infty}{3k choose k}dfrac{9k^2-3k-1}{(3k-1)(3k-2)}left(dfrac{2}{27}right)^k=dfrac{1}{4}
$$
After some simplification, I got the following result:
$$
sum_{k=0}^{infty}left{{3k choose k}+dfrac{27}{2k}{3k-4 choose 2k-3}right}left(dfrac{2}{27}right)^k.
$$
Now, how I can proceed?
real-analysis summation
real-analysis summation
edited Jan 20 at 18:27
Blue
48.7k870156
asked Jan 20 at 11:52
J.DoeJ.Doe
311
edited Jan 20 at 18:27
Blue
48.7k870156
asked Jan 20 at 11:52
J.DoeJ.Doe
311
edited Jan 20 at 18:27
Blue
48.7k870156
edited Jan 20 at 18:27
Blue
48.7k870156
edited Jan 20 at 18:27
Blue
48.7k870156
48.7k870156
asked Jan 20 at 11:52
J.DoeJ.Doe
311
asked Jan 20 at 11:52
J.DoeJ.Doe
311
asked Jan 20 at 11:52
J.DoeJ.Doe
311
311
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Starting with
$$
eqalign{
& S = sumlimits_{k = 0}^infty {
binom{3k}{k}{{9k^2 - 3k - 1} over {left( {3k - 1} right)left( {3k - 2} right)}}left( {{2 over {27}}} right)^{,k} } = cr
& = sumlimits_{k = 0}^infty {
binom{3k}{k} left( {1 + {1 over {left( {3k - 1} right)}} + {1 over {left( {3k - 2} right)}}} right)left( {{2 over {27}}} right)^{,k} } cr}
$$
we have that
$$
binom{3k}{k} = {{Gamma left( {3k + 1} right)} over {Gamma left( {2k + 1} right)}}{1 over {k!}}
$$
Using the $n$-plication formula for the Gamma function
$$
Gamma left( {n,z + 1} right) = Gamma left( {n,left( {z + 1/n} right)} right)quad =
{{n^{,n,z + 1/2} } over {left( {2,pi } right)^{left( {n - 1} right)/2} }}prodlimits_{1, le ,j, le ,n} {Gamma left( {z + {j over n}} right)}
$$
we get
$$
eqalign{
& {{Gamma left( {3,k + 1} right)} over {Gamma left( {2,k + 1} right)}}quad
= {{3^{,3,k + 1/2} } over {2^{,2,k + 1/2} sqrt {2,pi } ^, }}{{Gamma left( {k + {1 over 3}} right)Gamma left( {k + {2 over 3}} right)
Gamma left( {k + 1} right)} over {Gamma left( {k + {1 over 2}} right)Gamma left( {k + 1} right)}} = cr
& = sqrt {{3 over {4pi }}} {{Gamma left( {k + {1 over 3}} right)Gamma left( {k + {2 over 3}} right)}
over {Gamma left( {k + {1 over 2}} right)}}left( {{27 over 4}} right)^{,k} cr
& {{Gamma left( {3,k + 1} right)} over {Gamma left( {2,k + 1} right)left( {3k - 1} right)}}
= sqrt {{3 over {4pi }}} {{Gamma left( {k + {1 over 3}} right)Gamma left( {k + {2 over 3}} right)}
over {3Gamma left( {k + {1 over 2}} right)left( {k - 1/3} right)}}left( {{27 over 4}} right)^{,k} = cr
& = sqrt {{3 over {4pi }}} {1 over 3}{{Gamma left( {k + {1 over 3}} right)Gamma left( {k - {1 over 3}} right)}
over {Gamma left( {k + {1 over 2}} right)}}left( {{27 over 4}} right)^{,k} cr
& {{Gamma left( {3,k + 1} right)} over {Gamma left( {2,k + 1} right)left( {3k - 2} right)}}
= sqrt {{3 over {4pi }}} {{Gamma left( {k + {1 over 3}} right)Gamma left( {k + {2 over 3}} right)}
over {3Gamma left( {k + {1 over 2}} right)left( {k - 2/3} right)}}left( {{27 over 4}} right)^{,k} = cr
& = sqrt {{3 over {4pi }}} {1 over 3}{{Gamma left( {k - {2 over 3}} right)Gamma left( {k + {2 over 3}} right)}
over {Gamma left( {k + {1 over 2}} right)}}left( {{27 over 4}} right)^{,k} cr}
$$
So
$$
eqalign{
& S = cr
& = sumlimits_{k = 0}^infty {binom{3k}{k}
left( {1 + {1 over {left( {3k - 1} right)}} + {1 over {left( {3k - 2} right)}}} right)left( {{2 over {27}}} right)^{,k} } = cr
& = {1 over 2}sqrt {{1 over {3pi }}} sumlimits_{k = 0}^infty {
left( {3{{Gamma left( {k + {1 over 3}} right)Gamma left( {k + {2 over 3}} right)} over {Gamma left( {k + {1 over 2}} right)}}
+ {{Gamma left( {k + {1 over 3}} right)Gamma left( {k - {1 over 3}} right)} over {Gamma left( {k + {1 over 2}} right)}}
+ {{Gamma left( {k - {2 over 3}} right)Gamma left( {k + {2 over 3}} right)}
over {Gamma left( {k + {1 over 2}} right)}}} right){{left( {1/2} right)^{,k} } over {k!}}} cr}
$$
Using the Hypergeometric function that becomes
$$
eqalign{
& S = cr
& = {1 over 2}sqrt {{1 over {3pi }}} left( matrix{
3{{Gamma left( {1/3} right)Gamma left( {2/3} right)} over {Gamma left( {1/2} right)}}{}_2F_1 left( {left. {matrix{
{1/3,;2/3} cr {1/2} cr
} ,} right|;{1 over 2}} right) + hfill cr
+ {{Gamma left( {1/3} right)Gamma left( { - 1/3} right)} over {Gamma left( {1/2} right)}}{}_2F_1 left( {left. {matrix{
{1/3,; - 1/3} cr {1/2} cr
} ,} right|;{1 over 2}} right) + hfill cr
+ {{Gamma left( {2/3} right)Gamma left( { - 2/3} right)} over {Gamma left( {1/2} right)}}{}_2F_1 left( {left. {matrix{
{2/3,; - 2/3} cr {1/2} cr
} ,} right|;{1 over 2}} right) hfill cr} right) = cr
& = {}_2F_1 left( {left. {matrix{ {1/3,;2/3} cr {1/2} cr } ,} right|;{1 over 2}} right)
- {}_2F_1 left( {left. {matrix{ {1/3,; - 1/3} cr {1/2} cr } ,} right|;{1 over 2}} right)
- {1 over 2}{}_2F_1 left( {left. {matrix{ {2/3,; - 2/3} cr {1/2} cr } ,} right|;{1 over 2}} right) cr}
$$
Since the hypergeometric for the variable $z=1/2$
follows the following formulas (see e.g. this link)
$$
eqalign{
& {}_2F_1 left( {left. {matrix{ {a,;b} cr {{{a + b} over 2}} cr } ,} right|;{1 over 2}} right)
= sqrt pi ;Gamma left( {{{a + b} over 2}} right)left( {{1 over {Gamma left( {{{a + 1} over 2}} right)Gamma left( {{b over 2}} right)}}
+ {1 over {Gamma left( {{a over 2}} right)Gamma left( {{{b + 1} over 2}} right)}}} right) cr
& {}_2F_1 left( {left. {matrix{ {a,;b} cr {{{a + b + 1} over 2}} cr } ,} right|;{1 over 2}} right)
= sqrt pi ;Gamma left( {{{a + b + 1} over 2}} right){1 over {Gamma left( {{{a + 1} over 2}} right)Gamma left( {{{b + 1} over 2}} right)}} cr}
$$
then
$$
eqalign{
& {}_2F_1 left( {left. {matrix{ {1/3,;2/3} cr {1/2} cr } ,} right|;{1 over 2}} right)
= pi left( {{1 over {Gamma left( {2/3} right)Gamma left( {1/3} right)}}
+ {1 over {Gamma left( {1/6} right)Gamma left( {1 - 1/6} right)}}} right) = cr
& = left( {{3 over {2sqrt 3 }} + {1 over 2}} right) = {{sqrt 3 + 1} over 2} cr
& {}_2F_1 left( {left. {matrix{ {1/3,; - 1/3} cr {1/2} cr } ,} right|;{1 over 2}} right)
= pi {1 over {Gamma left( {2/3} right)Gamma left( {1/3} right)}} = {{sqrt 3 } over 2} cr
& {}_2F_1 left( {left. {matrix{ {2/3,; - 2/3} cr {1/2} cr } ,} right|;{1 over 2}} right)
= pi {1 over {Gamma left( {1 - {1 over 6}} right)Gamma left( {{1 over 6}} right)}} = {1 over 2} cr}
$$
And $S= 1/4$ follows.
answered Jan 20 at 18:24
G CabG Cab
19.8k31340
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add a comment |
$begingroup$
By Lagrange's inversion theorem (see Bring radical for the quintic analogue)
$$ sum_{kgeq 0}binom{3k}{k}left(frac{4}{27}right)^k x^{2k} = frac{cosleft(frac{1}{3}arcsin xright)}{sqrt{1-x^2}} tag{1}$$
and by partial fraction decomposition
$$ frac{9k^2-3k-1}{(3k-1)(3k-2)} = 1+frac{1}{3k-1}+frac{1}{3k-2}tag{2} $$
hence by evaluating $(1)$ at $x=frac{1}{sqrt{2}}$ we immediately get
$$ sum_{kgeq 0}binom{3k}{k}left(frac{2}{27}right)^k = frac{1+sqrt{3}}{2}. tag{3}$$
By reindexing the original series equals
$$-frac{1}{2}+sum_{kgeq 0}binom{3k}{k}frac{3(3k+2)(3k+1)}{(2k+2)(2k+1)}left[1+frac{1}{3k+1}+frac{1}{3k+2}right]left(frac{2}{27}right)^k tag{4}$$
hence it is enough to compute the integrals over $(0,1/sqrt{2})$ of the RHS of $(1)$ and of the RHS of $(1)$ multiplied by $x$. They are $frac{3}{2sqrt{2}}(sqrt{3}-1)$ and $frac{3}{16}(5-2sqrt{3})$ by the substitution $xmapstosintheta$. By re-combining these pieces the claim is proved.
answered Jan 20 at 18:35
Jack D'AurizioJack D'Aurizio
290k33284666
$endgroup$
1
$begingroup$
@J.Doe: An application of the Lagrange inversion theorem to calculate $sum_{k}binom{3k}{k}x^k$ can be found for instance at this post.
$endgroup$
– Markus Scheuer
Jan 20 at 19:21
add a comment |
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2 Answers
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2 Answers
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$begingroup$
Starting with
$$
eqalign{
& S = sumlimits_{k = 0}^infty {
binom{3k}{k}{{9k^2 - 3k - 1} over {left( {3k - 1} right)left( {3k - 2} right)}}left( {{2 over {27}}} right)^{,k} } = cr
& = sumlimits_{k = 0}^infty {
binom{3k}{k} left( {1 + {1 over {left( {3k - 1} right)}} + {1 over {left( {3k - 2} right)}}} right)left( {{2 over {27}}} right)^{,k} } cr}
$$
we have that
$$
binom{3k}{k} = {{Gamma left( {3k + 1} right)} over {Gamma left( {2k + 1} right)}}{1 over {k!}}
$$
Using the $n$-plication formula for the Gamma function
$$
Gamma left( {n,z + 1} right) = Gamma left( {n,left( {z + 1/n} right)} right)quad =
{{n^{,n,z + 1/2} } over {left( {2,pi } right)^{left( {n - 1} right)/2} }}prodlimits_{1, le ,j, le ,n} {Gamma left( {z + {j over n}} right)}
$$
we get
$$
eqalign{
& {{Gamma left( {3,k + 1} right)} over {Gamma left( {2,k + 1} right)}}quad
= {{3^{,3,k + 1/2} } over {2^{,2,k + 1/2} sqrt {2,pi } ^, }}{{Gamma left( {k + {1 over 3}} right)Gamma left( {k + {2 over 3}} right)
Gamma left( {k + 1} right)} over {Gamma left( {k + {1 over 2}} right)Gamma left( {k + 1} right)}} = cr
& = sqrt {{3 over {4pi }}} {{Gamma left( {k + {1 over 3}} right)Gamma left( {k + {2 over 3}} right)}
over {Gamma left( {k + {1 over 2}} right)}}left( {{27 over 4}} right)^{,k} cr
& {{Gamma left( {3,k + 1} right)} over {Gamma left( {2,k + 1} right)left( {3k - 1} right)}}
= sqrt {{3 over {4pi }}} {{Gamma left( {k + {1 over 3}} right)Gamma left( {k + {2 over 3}} right)}
over {3Gamma left( {k + {1 over 2}} right)left( {k - 1/3} right)}}left( {{27 over 4}} right)^{,k} = cr
& = sqrt {{3 over {4pi }}} {1 over 3}{{Gamma left( {k + {1 over 3}} right)Gamma left( {k - {1 over 3}} right)}
over {Gamma left( {k + {1 over 2}} right)}}left( {{27 over 4}} right)^{,k} cr
& {{Gamma left( {3,k + 1} right)} over {Gamma left( {2,k + 1} right)left( {3k - 2} right)}}
= sqrt {{3 over {4pi }}} {{Gamma left( {k + {1 over 3}} right)Gamma left( {k + {2 over 3}} right)}
over {3Gamma left( {k + {1 over 2}} right)left( {k - 2/3} right)}}left( {{27 over 4}} right)^{,k} = cr
& = sqrt {{3 over {4pi }}} {1 over 3}{{Gamma left( {k - {2 over 3}} right)Gamma left( {k + {2 over 3}} right)}
over {Gamma left( {k + {1 over 2}} right)}}left( {{27 over 4}} right)^{,k} cr}
$$
So
$$
eqalign{
& S = cr
& = sumlimits_{k = 0}^infty {binom{3k}{k}
left( {1 + {1 over {left( {3k - 1} right)}} + {1 over {left( {3k - 2} right)}}} right)left( {{2 over {27}}} right)^{,k} } = cr
& = {1 over 2}sqrt {{1 over {3pi }}} sumlimits_{k = 0}^infty {
left( {3{{Gamma left( {k + {1 over 3}} right)Gamma left( {k + {2 over 3}} right)} over {Gamma left( {k + {1 over 2}} right)}}
+ {{Gamma left( {k + {1 over 3}} right)Gamma left( {k - {1 over 3}} right)} over {Gamma left( {k + {1 over 2}} right)}}
+ {{Gamma left( {k - {2 over 3}} right)Gamma left( {k + {2 over 3}} right)}
over {Gamma left( {k + {1 over 2}} right)}}} right){{left( {1/2} right)^{,k} } over {k!}}} cr}
$$
Using the Hypergeometric function that becomes
$$
eqalign{
& S = cr
& = {1 over 2}sqrt {{1 over {3pi }}} left( matrix{
3{{Gamma left( {1/3} right)Gamma left( {2/3} right)} over {Gamma left( {1/2} right)}}{}_2F_1 left( {left. {matrix{
{1/3,;2/3} cr {1/2} cr
} ,} right|;{1 over 2}} right) + hfill cr
+ {{Gamma left( {1/3} right)Gamma left( { - 1/3} right)} over {Gamma left( {1/2} right)}}{}_2F_1 left( {left. {matrix{
{1/3,; - 1/3} cr {1/2} cr
} ,} right|;{1 over 2}} right) + hfill cr
+ {{Gamma left( {2/3} right)Gamma left( { - 2/3} right)} over {Gamma left( {1/2} right)}}{}_2F_1 left( {left. {matrix{
{2/3,; - 2/3} cr {1/2} cr
} ,} right|;{1 over 2}} right) hfill cr} right) = cr
& = {}_2F_1 left( {left. {matrix{ {1/3,;2/3} cr {1/2} cr } ,} right|;{1 over 2}} right)
- {}_2F_1 left( {left. {matrix{ {1/3,; - 1/3} cr {1/2} cr } ,} right|;{1 over 2}} right)
- {1 over 2}{}_2F_1 left( {left. {matrix{ {2/3,; - 2/3} cr {1/2} cr } ,} right|;{1 over 2}} right) cr}
$$
Since the hypergeometric for the variable $z=1/2$
follows the following formulas (see e.g. this link)
$$
eqalign{
& {}_2F_1 left( {left. {matrix{ {a,;b} cr {{{a + b} over 2}} cr } ,} right|;{1 over 2}} right)
= sqrt pi ;Gamma left( {{{a + b} over 2}} right)left( {{1 over {Gamma left( {{{a + 1} over 2}} right)Gamma left( {{b over 2}} right)}}
+ {1 over {Gamma left( {{a over 2}} right)Gamma left( {{{b + 1} over 2}} right)}}} right) cr
& {}_2F_1 left( {left. {matrix{ {a,;b} cr {{{a + b + 1} over 2}} cr } ,} right|;{1 over 2}} right)
= sqrt pi ;Gamma left( {{{a + b + 1} over 2}} right){1 over {Gamma left( {{{a + 1} over 2}} right)Gamma left( {{{b + 1} over 2}} right)}} cr}
$$
then
$$
eqalign{
& {}_2F_1 left( {left. {matrix{ {1/3,;2/3} cr {1/2} cr } ,} right|;{1 over 2}} right)
= pi left( {{1 over {Gamma left( {2/3} right)Gamma left( {1/3} right)}}
+ {1 over {Gamma left( {1/6} right)Gamma left( {1 - 1/6} right)}}} right) = cr
& = left( {{3 over {2sqrt 3 }} + {1 over 2}} right) = {{sqrt 3 + 1} over 2} cr
& {}_2F_1 left( {left. {matrix{ {1/3,; - 1/3} cr {1/2} cr } ,} right|;{1 over 2}} right)
= pi {1 over {Gamma left( {2/3} right)Gamma left( {1/3} right)}} = {{sqrt 3 } over 2} cr
& {}_2F_1 left( {left. {matrix{ {2/3,; - 2/3} cr {1/2} cr } ,} right|;{1 over 2}} right)
= pi {1 over {Gamma left( {1 - {1 over 6}} right)Gamma left( {{1 over 6}} right)}} = {1 over 2} cr}
$$
And $S= 1/4$ follows.
answered Jan 20 at 18:24
G CabG Cab
19.8k31340
$endgroup$
add a comment |
$begingroup$
Starting with
$$
eqalign{
& S = sumlimits_{k = 0}^infty {
binom{3k}{k}{{9k^2 - 3k - 1} over {left( {3k - 1} right)left( {3k - 2} right)}}left( {{2 over {27}}} right)^{,k} } = cr
& = sumlimits_{k = 0}^infty {
binom{3k}{k} left( {1 + {1 over {left( {3k - 1} right)}} + {1 over {left( {3k - 2} right)}}} right)left( {{2 over {27}}} right)^{,k} } cr}
$$
we have that
$$
binom{3k}{k} = {{Gamma left( {3k + 1} right)} over {Gamma left( {2k + 1} right)}}{1 over {k!}}
$$
Using the $n$-plication formula for the Gamma function
$$
Gamma left( {n,z + 1} right) = Gamma left( {n,left( {z + 1/n} right)} right)quad =
{{n^{,n,z + 1/2} } over {left( {2,pi } right)^{left( {n - 1} right)/2} }}prodlimits_{1, le ,j, le ,n} {Gamma left( {z + {j over n}} right)}
$$
we get
$$
eqalign{
& {{Gamma left( {3,k + 1} right)} over {Gamma left( {2,k + 1} right)}}quad
= {{3^{,3,k + 1/2} } over {2^{,2,k + 1/2} sqrt {2,pi } ^, }}{{Gamma left( {k + {1 over 3}} right)Gamma left( {k + {2 over 3}} right)
Gamma left( {k + 1} right)} over {Gamma left( {k + {1 over 2}} right)Gamma left( {k + 1} right)}} = cr
& = sqrt {{3 over {4pi }}} {{Gamma left( {k + {1 over 3}} right)Gamma left( {k + {2 over 3}} right)}
over {Gamma left( {k + {1 over 2}} right)}}left( {{27 over 4}} right)^{,k} cr
& {{Gamma left( {3,k + 1} right)} over {Gamma left( {2,k + 1} right)left( {3k - 1} right)}}
= sqrt {{3 over {4pi }}} {{Gamma left( {k + {1 over 3}} right)Gamma left( {k + {2 over 3}} right)}
over {3Gamma left( {k + {1 over 2}} right)left( {k - 1/3} right)}}left( {{27 over 4}} right)^{,k} = cr
& = sqrt {{3 over {4pi }}} {1 over 3}{{Gamma left( {k + {1 over 3}} right)Gamma left( {k - {1 over 3}} right)}
over {Gamma left( {k + {1 over 2}} right)}}left( {{27 over 4}} right)^{,k} cr
& {{Gamma left( {3,k + 1} right)} over {Gamma left( {2,k + 1} right)left( {3k - 2} right)}}
= sqrt {{3 over {4pi }}} {{Gamma left( {k + {1 over 3}} right)Gamma left( {k + {2 over 3}} right)}
over {3Gamma left( {k + {1 over 2}} right)left( {k - 2/3} right)}}left( {{27 over 4}} right)^{,k} = cr
& = sqrt {{3 over {4pi }}} {1 over 3}{{Gamma left( {k - {2 over 3}} right)Gamma left( {k + {2 over 3}} right)}
over {Gamma left( {k + {1 over 2}} right)}}left( {{27 over 4}} right)^{,k} cr}
$$
So
$$
eqalign{
& S = cr
& = sumlimits_{k = 0}^infty {binom{3k}{k}
left( {1 + {1 over {left( {3k - 1} right)}} + {1 over {left( {3k - 2} right)}}} right)left( {{2 over {27}}} right)^{,k} } = cr
& = {1 over 2}sqrt {{1 over {3pi }}} sumlimits_{k = 0}^infty {
left( {3{{Gamma left( {k + {1 over 3}} right)Gamma left( {k + {2 over 3}} right)} over {Gamma left( {k + {1 over 2}} right)}}
+ {{Gamma left( {k + {1 over 3}} right)Gamma left( {k - {1 over 3}} right)} over {Gamma left( {k + {1 over 2}} right)}}
+ {{Gamma left( {k - {2 over 3}} right)Gamma left( {k + {2 over 3}} right)}
over {Gamma left( {k + {1 over 2}} right)}}} right){{left( {1/2} right)^{,k} } over {k!}}} cr}
$$
Using the Hypergeometric function that becomes
$$
eqalign{
& S = cr
& = {1 over 2}sqrt {{1 over {3pi }}} left( matrix{
3{{Gamma left( {1/3} right)Gamma left( {2/3} right)} over {Gamma left( {1/2} right)}}{}_2F_1 left( {left. {matrix{
{1/3,;2/3} cr {1/2} cr
} ,} right|;{1 over 2}} right) + hfill cr
+ {{Gamma left( {1/3} right)Gamma left( { - 1/3} right)} over {Gamma left( {1/2} right)}}{}_2F_1 left( {left. {matrix{
{1/3,; - 1/3} cr {1/2} cr
} ,} right|;{1 over 2}} right) + hfill cr
+ {{Gamma left( {2/3} right)Gamma left( { - 2/3} right)} over {Gamma left( {1/2} right)}}{}_2F_1 left( {left. {matrix{
{2/3,; - 2/3} cr {1/2} cr
} ,} right|;{1 over 2}} right) hfill cr} right) = cr
& = {}_2F_1 left( {left. {matrix{ {1/3,;2/3} cr {1/2} cr } ,} right|;{1 over 2}} right)
- {}_2F_1 left( {left. {matrix{ {1/3,; - 1/3} cr {1/2} cr } ,} right|;{1 over 2}} right)
- {1 over 2}{}_2F_1 left( {left. {matrix{ {2/3,; - 2/3} cr {1/2} cr } ,} right|;{1 over 2}} right) cr}
$$
Since the hypergeometric for the variable $z=1/2$
follows the following formulas (see e.g. this link)
$$
eqalign{
& {}_2F_1 left( {left. {matrix{ {a,;b} cr {{{a + b} over 2}} cr } ,} right|;{1 over 2}} right)
= sqrt pi ;Gamma left( {{{a + b} over 2}} right)left( {{1 over {Gamma left( {{{a + 1} over 2}} right)Gamma left( {{b over 2}} right)}}
+ {1 over {Gamma left( {{a over 2}} right)Gamma left( {{{b + 1} over 2}} right)}}} right) cr
& {}_2F_1 left( {left. {matrix{ {a,;b} cr {{{a + b + 1} over 2}} cr } ,} right|;{1 over 2}} right)
= sqrt pi ;Gamma left( {{{a + b + 1} over 2}} right){1 over {Gamma left( {{{a + 1} over 2}} right)Gamma left( {{{b + 1} over 2}} right)}} cr}
$$
then
$$
eqalign{
& {}_2F_1 left( {left. {matrix{ {1/3,;2/3} cr {1/2} cr } ,} right|;{1 over 2}} right)
= pi left( {{1 over {Gamma left( {2/3} right)Gamma left( {1/3} right)}}
+ {1 over {Gamma left( {1/6} right)Gamma left( {1 - 1/6} right)}}} right) = cr
& = left( {{3 over {2sqrt 3 }} + {1 over 2}} right) = {{sqrt 3 + 1} over 2} cr
& {}_2F_1 left( {left. {matrix{ {1/3,; - 1/3} cr {1/2} cr } ,} right|;{1 over 2}} right)
= pi {1 over {Gamma left( {2/3} right)Gamma left( {1/3} right)}} = {{sqrt 3 } over 2} cr
& {}_2F_1 left( {left. {matrix{ {2/3,; - 2/3} cr {1/2} cr } ,} right|;{1 over 2}} right)
= pi {1 over {Gamma left( {1 - {1 over 6}} right)Gamma left( {{1 over 6}} right)}} = {1 over 2} cr}
$$
And $S= 1/4$ follows.
answered Jan 20 at 18:24
G CabG Cab
19.8k31340
$endgroup$
add a comment |
$begingroup$
Starting with
$$
eqalign{
& S = sumlimits_{k = 0}^infty {
binom{3k}{k}{{9k^2 - 3k - 1} over {left( {3k - 1} right)left( {3k - 2} right)}}left( {{2 over {27}}} right)^{,k} } = cr
& = sumlimits_{k = 0}^infty {
binom{3k}{k} left( {1 + {1 over {left( {3k - 1} right)}} + {1 over {left( {3k - 2} right)}}} right)left( {{2 over {27}}} right)^{,k} } cr}
$$
we have that
$$
binom{3k}{k} = {{Gamma left( {3k + 1} right)} over {Gamma left( {2k + 1} right)}}{1 over {k!}}
$$
Using the $n$-plication formula for the Gamma function
$$
Gamma left( {n,z + 1} right) = Gamma left( {n,left( {z + 1/n} right)} right)quad =
{{n^{,n,z + 1/2} } over {left( {2,pi } right)^{left( {n - 1} right)/2} }}prodlimits_{1, le ,j, le ,n} {Gamma left( {z + {j over n}} right)}
$$
we get
$$
eqalign{
& {{Gamma left( {3,k + 1} right)} over {Gamma left( {2,k + 1} right)}}quad
= {{3^{,3,k + 1/2} } over {2^{,2,k + 1/2} sqrt {2,pi } ^, }}{{Gamma left( {k + {1 over 3}} right)Gamma left( {k + {2 over 3}} right)
Gamma left( {k + 1} right)} over {Gamma left( {k + {1 over 2}} right)Gamma left( {k + 1} right)}} = cr
& = sqrt {{3 over {4pi }}} {{Gamma left( {k + {1 over 3}} right)Gamma left( {k + {2 over 3}} right)}
over {Gamma left( {k + {1 over 2}} right)}}left( {{27 over 4}} right)^{,k} cr
& {{Gamma left( {3,k + 1} right)} over {Gamma left( {2,k + 1} right)left( {3k - 1} right)}}
= sqrt {{3 over {4pi }}} {{Gamma left( {k + {1 over 3}} right)Gamma left( {k + {2 over 3}} right)}
over {3Gamma left( {k + {1 over 2}} right)left( {k - 1/3} right)}}left( {{27 over 4}} right)^{,k} = cr
& = sqrt {{3 over {4pi }}} {1 over 3}{{Gamma left( {k + {1 over 3}} right)Gamma left( {k - {1 over 3}} right)}
over {Gamma left( {k + {1 over 2}} right)}}left( {{27 over 4}} right)^{,k} cr
& {{Gamma left( {3,k + 1} right)} over {Gamma left( {2,k + 1} right)left( {3k - 2} right)}}
= sqrt {{3 over {4pi }}} {{Gamma left( {k + {1 over 3}} right)Gamma left( {k + {2 over 3}} right)}
over {3Gamma left( {k + {1 over 2}} right)left( {k - 2/3} right)}}left( {{27 over 4}} right)^{,k} = cr
& = sqrt {{3 over {4pi }}} {1 over 3}{{Gamma left( {k - {2 over 3}} right)Gamma left( {k + {2 over 3}} right)}
over {Gamma left( {k + {1 over 2}} right)}}left( {{27 over 4}} right)^{,k} cr}
$$
So
$$
eqalign{
& S = cr
& = sumlimits_{k = 0}^infty {binom{3k}{k}
left( {1 + {1 over {left( {3k - 1} right)}} + {1 over {left( {3k - 2} right)}}} right)left( {{2 over {27}}} right)^{,k} } = cr
& = {1 over 2}sqrt {{1 over {3pi }}} sumlimits_{k = 0}^infty {
left( {3{{Gamma left( {k + {1 over 3}} right)Gamma left( {k + {2 over 3}} right)} over {Gamma left( {k + {1 over 2}} right)}}
+ {{Gamma left( {k + {1 over 3}} right)Gamma left( {k - {1 over 3}} right)} over {Gamma left( {k + {1 over 2}} right)}}
+ {{Gamma left( {k - {2 over 3}} right)Gamma left( {k + {2 over 3}} right)}
over {Gamma left( {k + {1 over 2}} right)}}} right){{left( {1/2} right)^{,k} } over {k!}}} cr}
$$
Using the Hypergeometric function that becomes
$$
eqalign{
& S = cr
& = {1 over 2}sqrt {{1 over {3pi }}} left( matrix{
3{{Gamma left( {1/3} right)Gamma left( {2/3} right)} over {Gamma left( {1/2} right)}}{}_2F_1 left( {left. {matrix{
{1/3,;2/3} cr {1/2} cr
} ,} right|;{1 over 2}} right) + hfill cr
+ {{Gamma left( {1/3} right)Gamma left( { - 1/3} right)} over {Gamma left( {1/2} right)}}{}_2F_1 left( {left. {matrix{
{1/3,; - 1/3} cr {1/2} cr
} ,} right|;{1 over 2}} right) + hfill cr
+ {{Gamma left( {2/3} right)Gamma left( { - 2/3} right)} over {Gamma left( {1/2} right)}}{}_2F_1 left( {left. {matrix{
{2/3,; - 2/3} cr {1/2} cr
} ,} right|;{1 over 2}} right) hfill cr} right) = cr
& = {}_2F_1 left( {left. {matrix{ {1/3,;2/3} cr {1/2} cr } ,} right|;{1 over 2}} right)
- {}_2F_1 left( {left. {matrix{ {1/3,; - 1/3} cr {1/2} cr } ,} right|;{1 over 2}} right)
- {1 over 2}{}_2F_1 left( {left. {matrix{ {2/3,; - 2/3} cr {1/2} cr } ,} right|;{1 over 2}} right) cr}
$$
Since the hypergeometric for the variable $z=1/2$
follows the following formulas (see e.g. this link)
$$
eqalign{
& {}_2F_1 left( {left. {matrix{ {a,;b} cr {{{a + b} over 2}} cr } ,} right|;{1 over 2}} right)
= sqrt pi ;Gamma left( {{{a + b} over 2}} right)left( {{1 over {Gamma left( {{{a + 1} over 2}} right)Gamma left( {{b over 2}} right)}}
+ {1 over {Gamma left( {{a over 2}} right)Gamma left( {{{b + 1} over 2}} right)}}} right) cr
& {}_2F_1 left( {left. {matrix{ {a,;b} cr {{{a + b + 1} over 2}} cr } ,} right|;{1 over 2}} right)
= sqrt pi ;Gamma left( {{{a + b + 1} over 2}} right){1 over {Gamma left( {{{a + 1} over 2}} right)Gamma left( {{{b + 1} over 2}} right)}} cr}
$$
then
$$
eqalign{
& {}_2F_1 left( {left. {matrix{ {1/3,;2/3} cr {1/2} cr } ,} right|;{1 over 2}} right)
= pi left( {{1 over {Gamma left( {2/3} right)Gamma left( {1/3} right)}}
+ {1 over {Gamma left( {1/6} right)Gamma left( {1 - 1/6} right)}}} right) = cr
& = left( {{3 over {2sqrt 3 }} + {1 over 2}} right) = {{sqrt 3 + 1} over 2} cr
& {}_2F_1 left( {left. {matrix{ {1/3,; - 1/3} cr {1/2} cr } ,} right|;{1 over 2}} right)
= pi {1 over {Gamma left( {2/3} right)Gamma left( {1/3} right)}} = {{sqrt 3 } over 2} cr
& {}_2F_1 left( {left. {matrix{ {2/3,; - 2/3} cr {1/2} cr } ,} right|;{1 over 2}} right)
= pi {1 over {Gamma left( {1 - {1 over 6}} right)Gamma left( {{1 over 6}} right)}} = {1 over 2} cr}
$$
And $S= 1/4$ follows.
answered Jan 20 at 18:24
G CabG Cab
19.8k31340
$endgroup$
Starting with
$$
eqalign{
& S = sumlimits_{k = 0}^infty {
binom{3k}{k}{{9k^2 - 3k - 1} over {left( {3k - 1} right)left( {3k - 2} right)}}left( {{2 over {27}}} right)^{,k} } = cr
& = sumlimits_{k = 0}^infty {
binom{3k}{k} left( {1 + {1 over {left( {3k - 1} right)}} + {1 over {left( {3k - 2} right)}}} right)left( {{2 over {27}}} right)^{,k} } cr}
$$
we have that
$$
binom{3k}{k} = {{Gamma left( {3k + 1} right)} over {Gamma left( {2k + 1} right)}}{1 over {k!}}
$$
Using the $n$-plication formula for the Gamma function
$$
Gamma left( {n,z + 1} right) = Gamma left( {n,left( {z + 1/n} right)} right)quad =
{{n^{,n,z + 1/2} } over {left( {2,pi } right)^{left( {n - 1} right)/2} }}prodlimits_{1, le ,j, le ,n} {Gamma left( {z + {j over n}} right)}
$$
we get
$$
eqalign{
& {{Gamma left( {3,k + 1} right)} over {Gamma left( {2,k + 1} right)}}quad
= {{3^{,3,k + 1/2} } over {2^{,2,k + 1/2} sqrt {2,pi } ^, }}{{Gamma left( {k + {1 over 3}} right)Gamma left( {k + {2 over 3}} right)
Gamma left( {k + 1} right)} over {Gamma left( {k + {1 over 2}} right)Gamma left( {k + 1} right)}} = cr
& = sqrt {{3 over {4pi }}} {{Gamma left( {k + {1 over 3}} right)Gamma left( {k + {2 over 3}} right)}
over {Gamma left( {k + {1 over 2}} right)}}left( {{27 over 4}} right)^{,k} cr
& {{Gamma left( {3,k + 1} right)} over {Gamma left( {2,k + 1} right)left( {3k - 1} right)}}
= sqrt {{3 over {4pi }}} {{Gamma left( {k + {1 over 3}} right)Gamma left( {k + {2 over 3}} right)}
over {3Gamma left( {k + {1 over 2}} right)left( {k - 1/3} right)}}left( {{27 over 4}} right)^{,k} = cr
& = sqrt {{3 over {4pi }}} {1 over 3}{{Gamma left( {k + {1 over 3}} right)Gamma left( {k - {1 over 3}} right)}
over {Gamma left( {k + {1 over 2}} right)}}left( {{27 over 4}} right)^{,k} cr
& {{Gamma left( {3,k + 1} right)} over {Gamma left( {2,k + 1} right)left( {3k - 2} right)}}
= sqrt {{3 over {4pi }}} {{Gamma left( {k + {1 over 3}} right)Gamma left( {k + {2 over 3}} right)}
over {3Gamma left( {k + {1 over 2}} right)left( {k - 2/3} right)}}left( {{27 over 4}} right)^{,k} = cr
& = sqrt {{3 over {4pi }}} {1 over 3}{{Gamma left( {k - {2 over 3}} right)Gamma left( {k + {2 over 3}} right)}
over {Gamma left( {k + {1 over 2}} right)}}left( {{27 over 4}} right)^{,k} cr}
$$
So
$$
eqalign{
& S = cr
& = sumlimits_{k = 0}^infty {binom{3k}{k}
left( {1 + {1 over {left( {3k - 1} right)}} + {1 over {left( {3k - 2} right)}}} right)left( {{2 over {27}}} right)^{,k} } = cr
& = {1 over 2}sqrt {{1 over {3pi }}} sumlimits_{k = 0}^infty {
left( {3{{Gamma left( {k + {1 over 3}} right)Gamma left( {k + {2 over 3}} right)} over {Gamma left( {k + {1 over 2}} right)}}
+ {{Gamma left( {k + {1 over 3}} right)Gamma left( {k - {1 over 3}} right)} over {Gamma left( {k + {1 over 2}} right)}}
+ {{Gamma left( {k - {2 over 3}} right)Gamma left( {k + {2 over 3}} right)}
over {Gamma left( {k + {1 over 2}} right)}}} right){{left( {1/2} right)^{,k} } over {k!}}} cr}
$$
Using the Hypergeometric function that becomes
$$
eqalign{
& S = cr
& = {1 over 2}sqrt {{1 over {3pi }}} left( matrix{
3{{Gamma left( {1/3} right)Gamma left( {2/3} right)} over {Gamma left( {1/2} right)}}{}_2F_1 left( {left. {matrix{
{1/3,;2/3} cr {1/2} cr
} ,} right|;{1 over 2}} right) + hfill cr
+ {{Gamma left( {1/3} right)Gamma left( { - 1/3} right)} over {Gamma left( {1/2} right)}}{}_2F_1 left( {left. {matrix{
{1/3,; - 1/3} cr {1/2} cr
} ,} right|;{1 over 2}} right) + hfill cr
+ {{Gamma left( {2/3} right)Gamma left( { - 2/3} right)} over {Gamma left( {1/2} right)}}{}_2F_1 left( {left. {matrix{
{2/3,; - 2/3} cr {1/2} cr
} ,} right|;{1 over 2}} right) hfill cr} right) = cr
& = {}_2F_1 left( {left. {matrix{ {1/3,;2/3} cr {1/2} cr } ,} right|;{1 over 2}} right)
- {}_2F_1 left( {left. {matrix{ {1/3,; - 1/3} cr {1/2} cr } ,} right|;{1 over 2}} right)
- {1 over 2}{}_2F_1 left( {left. {matrix{ {2/3,; - 2/3} cr {1/2} cr } ,} right|;{1 over 2}} right) cr}
$$
Since the hypergeometric for the variable $z=1/2$
follows the following formulas (see e.g. this link)
$$
eqalign{
& {}_2F_1 left( {left. {matrix{ {a,;b} cr {{{a + b} over 2}} cr } ,} right|;{1 over 2}} right)
= sqrt pi ;Gamma left( {{{a + b} over 2}} right)left( {{1 over {Gamma left( {{{a + 1} over 2}} right)Gamma left( {{b over 2}} right)}}
+ {1 over {Gamma left( {{a over 2}} right)Gamma left( {{{b + 1} over 2}} right)}}} right) cr
& {}_2F_1 left( {left. {matrix{ {a,;b} cr {{{a + b + 1} over 2}} cr } ,} right|;{1 over 2}} right)
= sqrt pi ;Gamma left( {{{a + b + 1} over 2}} right){1 over {Gamma left( {{{a + 1} over 2}} right)Gamma left( {{{b + 1} over 2}} right)}} cr}
$$
then
$$
eqalign{
& {}_2F_1 left( {left. {matrix{ {1/3,;2/3} cr {1/2} cr } ,} right|;{1 over 2}} right)
= pi left( {{1 over {Gamma left( {2/3} right)Gamma left( {1/3} right)}}
+ {1 over {Gamma left( {1/6} right)Gamma left( {1 - 1/6} right)}}} right) = cr
& = left( {{3 over {2sqrt 3 }} + {1 over 2}} right) = {{sqrt 3 + 1} over 2} cr
& {}_2F_1 left( {left. {matrix{ {1/3,; - 1/3} cr {1/2} cr } ,} right|;{1 over 2}} right)
= pi {1 over {Gamma left( {2/3} right)Gamma left( {1/3} right)}} = {{sqrt 3 } over 2} cr
& {}_2F_1 left( {left. {matrix{ {2/3,; - 2/3} cr {1/2} cr } ,} right|;{1 over 2}} right)
= pi {1 over {Gamma left( {1 - {1 over 6}} right)Gamma left( {{1 over 6}} right)}} = {1 over 2} cr}
$$
And $S= 1/4$ follows.
answered Jan 20 at 18:24
G CabG Cab
19.8k31340
edited Jan 20 at 22:45
answered Jan 20 at 18:24
G CabG Cab
19.8k31340
answered Jan 20 at 18:24
G CabG Cab
19.8k31340
answered Jan 20 at 18:24
G CabG Cab
19.8k31340
19.8k31340
add a comment |
add a comment |
$begingroup$
By Lagrange's inversion theorem (see Bring radical for the quintic analogue)
$$ sum_{kgeq 0}binom{3k}{k}left(frac{4}{27}right)^k x^{2k} = frac{cosleft(frac{1}{3}arcsin xright)}{sqrt{1-x^2}} tag{1}$$
and by partial fraction decomposition
$$ frac{9k^2-3k-1}{(3k-1)(3k-2)} = 1+frac{1}{3k-1}+frac{1}{3k-2}tag{2} $$
hence by evaluating $(1)$ at $x=frac{1}{sqrt{2}}$ we immediately get
$$ sum_{kgeq 0}binom{3k}{k}left(frac{2}{27}right)^k = frac{1+sqrt{3}}{2}. tag{3}$$
By reindexing the original series equals
$$-frac{1}{2}+sum_{kgeq 0}binom{3k}{k}frac{3(3k+2)(3k+1)}{(2k+2)(2k+1)}left[1+frac{1}{3k+1}+frac{1}{3k+2}right]left(frac{2}{27}right)^k tag{4}$$
hence it is enough to compute the integrals over $(0,1/sqrt{2})$ of the RHS of $(1)$ and of the RHS of $(1)$ multiplied by $x$. They are $frac{3}{2sqrt{2}}(sqrt{3}-1)$ and $frac{3}{16}(5-2sqrt{3})$ by the substitution $xmapstosintheta$. By re-combining these pieces the claim is proved.
answered Jan 20 at 18:35
Jack D'AurizioJack D'Aurizio
290k33284666
$endgroup$
1
$begingroup$
@J.Doe: An application of the Lagrange inversion theorem to calculate $sum_{k}binom{3k}{k}x^k$ can be found for instance at this post.
$endgroup$
– Markus Scheuer
Jan 20 at 19:21
add a comment |
$begingroup$
By Lagrange's inversion theorem (see Bring radical for the quintic analogue)
$$ sum_{kgeq 0}binom{3k}{k}left(frac{4}{27}right)^k x^{2k} = frac{cosleft(frac{1}{3}arcsin xright)}{sqrt{1-x^2}} tag{1}$$
and by partial fraction decomposition
$$ frac{9k^2-3k-1}{(3k-1)(3k-2)} = 1+frac{1}{3k-1}+frac{1}{3k-2}tag{2} $$
hence by evaluating $(1)$ at $x=frac{1}{sqrt{2}}$ we immediately get
$$ sum_{kgeq 0}binom{3k}{k}left(frac{2}{27}right)^k = frac{1+sqrt{3}}{2}. tag{3}$$
By reindexing the original series equals
$$-frac{1}{2}+sum_{kgeq 0}binom{3k}{k}frac{3(3k+2)(3k+1)}{(2k+2)(2k+1)}left[1+frac{1}{3k+1}+frac{1}{3k+2}right]left(frac{2}{27}right)^k tag{4}$$
hence it is enough to compute the integrals over $(0,1/sqrt{2})$ of the RHS of $(1)$ and of the RHS of $(1)$ multiplied by $x$. They are $frac{3}{2sqrt{2}}(sqrt{3}-1)$ and $frac{3}{16}(5-2sqrt{3})$ by the substitution $xmapstosintheta$. By re-combining these pieces the claim is proved.
answered Jan 20 at 18:35
Jack D'AurizioJack D'Aurizio
290k33284666
$endgroup$
1
$begingroup$
@J.Doe: An application of the Lagrange inversion theorem to calculate $sum_{k}binom{3k}{k}x^k$ can be found for instance at this post.
$endgroup$
– Markus Scheuer
Jan 20 at 19:21
add a comment |
$begingroup$
By Lagrange's inversion theorem (see Bring radical for the quintic analogue)
$$ sum_{kgeq 0}binom{3k}{k}left(frac{4}{27}right)^k x^{2k} = frac{cosleft(frac{1}{3}arcsin xright)}{sqrt{1-x^2}} tag{1}$$
and by partial fraction decomposition
$$ frac{9k^2-3k-1}{(3k-1)(3k-2)} = 1+frac{1}{3k-1}+frac{1}{3k-2}tag{2} $$
hence by evaluating $(1)$ at $x=frac{1}{sqrt{2}}$ we immediately get
$$ sum_{kgeq 0}binom{3k}{k}left(frac{2}{27}right)^k = frac{1+sqrt{3}}{2}. tag{3}$$
By reindexing the original series equals
$$-frac{1}{2}+sum_{kgeq 0}binom{3k}{k}frac{3(3k+2)(3k+1)}{(2k+2)(2k+1)}left[1+frac{1}{3k+1}+frac{1}{3k+2}right]left(frac{2}{27}right)^k tag{4}$$
hence it is enough to compute the integrals over $(0,1/sqrt{2})$ of the RHS of $(1)$ and of the RHS of $(1)$ multiplied by $x$. They are $frac{3}{2sqrt{2}}(sqrt{3}-1)$ and $frac{3}{16}(5-2sqrt{3})$ by the substitution $xmapstosintheta$. By re-combining these pieces the claim is proved.
answered Jan 20 at 18:35
Jack D'AurizioJack D'Aurizio
290k33284666
$endgroup$
By Lagrange's inversion theorem (see Bring radical for the quintic analogue)
$$ sum_{kgeq 0}binom{3k}{k}left(frac{4}{27}right)^k x^{2k} = frac{cosleft(frac{1}{3}arcsin xright)}{sqrt{1-x^2}} tag{1}$$
and by partial fraction decomposition
$$ frac{9k^2-3k-1}{(3k-1)(3k-2)} = 1+frac{1}{3k-1}+frac{1}{3k-2}tag{2} $$
hence by evaluating $(1)$ at $x=frac{1}{sqrt{2}}$ we immediately get
$$ sum_{kgeq 0}binom{3k}{k}left(frac{2}{27}right)^k = frac{1+sqrt{3}}{2}. tag{3}$$
By reindexing the original series equals
$$-frac{1}{2}+sum_{kgeq 0}binom{3k}{k}frac{3(3k+2)(3k+1)}{(2k+2)(2k+1)}left[1+frac{1}{3k+1}+frac{1}{3k+2}right]left(frac{2}{27}right)^k tag{4}$$
hence it is enough to compute the integrals over $(0,1/sqrt{2})$ of the RHS of $(1)$ and of the RHS of $(1)$ multiplied by $x$. They are $frac{3}{2sqrt{2}}(sqrt{3}-1)$ and $frac{3}{16}(5-2sqrt{3})$ by the substitution $xmapstosintheta$. By re-combining these pieces the claim is proved.
answered Jan 20 at 18:35
Jack D'AurizioJack D'Aurizio
290k33284666
answered Jan 20 at 18:35
Jack D'AurizioJack D'Aurizio
290k33284666
answered Jan 20 at 18:35
Jack D'AurizioJack D'Aurizio
290k33284666
answered Jan 20 at 18:35
Jack D'AurizioJack D'Aurizio
290k33284666
290k33284666
1
$begingroup$
@J.Doe: An application of the Lagrange inversion theorem to calculate $sum_{k}binom{3k}{k}x^k$ can be found for instance at this post.
$endgroup$
– Markus Scheuer
Jan 20 at 19:21
add a comment |
1
$begingroup$
@J.Doe: An application of the Lagrange inversion theorem to calculate $sum_{k}binom{3k}{k}x^k$ can be found for instance at this post.
$endgroup$
– Markus Scheuer
Jan 20 at 19:21
1
1
$begingroup$
@J.Doe: An application of the Lagrange inversion theorem to calculate $sum_{k}binom{3k}{k}x^k$ can be found for instance at this post.
$endgroup$
– Markus Scheuer
Jan 20 at 19:21
$begingroup$
@J.Doe: An application of the Lagrange inversion theorem to calculate $sum_{k}binom{3k}{k}x^k$ can be found for instance at this post.
$endgroup$
– Markus Scheuer
Jan 20 at 19:21
add a comment |
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