Concerning this sum $sum_{k=1}^{infty}{2k choose k}^2 4^{-2k}cdot frac{1}{(ak)^3-ak}$
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I was looking at this Ramanujan phi-Function
Let:
$$R(a)=sum_{k=1}^{infty}frac{1}{(ak)^3-ak}tag1$$
and this paper of the form,
$$sum_{k=1}^{infty}frac{{2k choose k}}{4^k}tag2$$ and
$$sum_{k=1}^{infty}left[frac{{2k choose k}}{4^k}right]^2tag 3$$
By combining them together we have
$$S(a)=sum_{k=1}^{infty}frac{{2k choose k}}{4^k}cdot frac{1}{(ak)^3-ak}tag4$$ and
$$T(a)=sum_{k=1}^{infty}left[frac{{2k choose k}}{4^k}right]^2cdot frac{1}{(ak)^3-ak}tag5$$
The conjectured closed form for $(4)$ and $(5)$, where $a=2$ are
$$S(2)=sum_{k=1}^{infty}frac{{2k choose k}}{4^k}cdot frac{1}{(2k)^3-2k}=frac{pi}{4}-ln(2)tag6$$ and
$$T(2)=sum_{k=1}^{infty}left[frac{{2k choose k}}{4^k}right]^2cdot frac{1}{(2k)^3-2k}=frac{6G-piln(4)-1}{pi}tag7$$
Where G is the Catalan's constant.
How can we prove these conjectures?
sequences-and-series
$endgroup$
add a comment |
$begingroup$
I was looking at this Ramanujan phi-Function
Let:
$$R(a)=sum_{k=1}^{infty}frac{1}{(ak)^3-ak}tag1$$
and this paper of the form,
$$sum_{k=1}^{infty}frac{{2k choose k}}{4^k}tag2$$ and
$$sum_{k=1}^{infty}left[frac{{2k choose k}}{4^k}right]^2tag 3$$
By combining them together we have
$$S(a)=sum_{k=1}^{infty}frac{{2k choose k}}{4^k}cdot frac{1}{(ak)^3-ak}tag4$$ and
$$T(a)=sum_{k=1}^{infty}left[frac{{2k choose k}}{4^k}right]^2cdot frac{1}{(ak)^3-ak}tag5$$
The conjectured closed form for $(4)$ and $(5)$, where $a=2$ are
$$S(2)=sum_{k=1}^{infty}frac{{2k choose k}}{4^k}cdot frac{1}{(2k)^3-2k}=frac{pi}{4}-ln(2)tag6$$ and
$$T(2)=sum_{k=1}^{infty}left[frac{{2k choose k}}{4^k}right]^2cdot frac{1}{(2k)^3-2k}=frac{6G-piln(4)-1}{pi}tag7$$
Where G is the Catalan's constant.
How can we prove these conjectures?
sequences-and-series
$endgroup$
$begingroup$
Is it ${2n choose n}^2 4^{-2n}$ or ${2k choose k}^2 4^{-2k}$ in the headline?
$endgroup$
– Andreas
Jan 3 at 9:43
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Why don't you try the methods of the paper? ;)
$endgroup$
– Jack D'Aurizio
Jan 3 at 17:04
$begingroup$
Please notice that $(2)$ is actually a divergent series, since $frac{1}{4^n}binom{2n}{n}simfrac{1}{sqrt{pi n}}$, so "this paper of the form divergent series" does not make much sense. Better to say "this paper about the weight $frac{1}{4^n}binom{2n}{n}$ and its square".
$endgroup$
– Jack D'Aurizio
Jan 3 at 17:20
$begingroup$
$(3)$ is also a divergent series (same reason), while $$sum_{ngeq 0}left[frac{1}{4^n}binom{2n}{n}right]^3$$ is related to the lemniscate constant (hence to $Gammaleft(frac{1}{4}right)$) by many good reasons; among them, Clausen's formula for the square of particular $phantom{}_2 F_1$s.
$endgroup$
– Jack D'Aurizio
Jan 3 at 17:32
$begingroup$
$$sum_{ngeq 0}left[frac{1}{4^n}binom{2n}{n}right]^4$$ is much more obscure: see math.stackexchange.com/questions/2506266/…
$endgroup$
– Jack D'Aurizio
Jan 3 at 17:34
add a comment |
$begingroup$
I was looking at this Ramanujan phi-Function
Let:
$$R(a)=sum_{k=1}^{infty}frac{1}{(ak)^3-ak}tag1$$
and this paper of the form,
$$sum_{k=1}^{infty}frac{{2k choose k}}{4^k}tag2$$ and
$$sum_{k=1}^{infty}left[frac{{2k choose k}}{4^k}right]^2tag 3$$
By combining them together we have
$$S(a)=sum_{k=1}^{infty}frac{{2k choose k}}{4^k}cdot frac{1}{(ak)^3-ak}tag4$$ and
$$T(a)=sum_{k=1}^{infty}left[frac{{2k choose k}}{4^k}right]^2cdot frac{1}{(ak)^3-ak}tag5$$
The conjectured closed form for $(4)$ and $(5)$, where $a=2$ are
$$S(2)=sum_{k=1}^{infty}frac{{2k choose k}}{4^k}cdot frac{1}{(2k)^3-2k}=frac{pi}{4}-ln(2)tag6$$ and
$$T(2)=sum_{k=1}^{infty}left[frac{{2k choose k}}{4^k}right]^2cdot frac{1}{(2k)^3-2k}=frac{6G-piln(4)-1}{pi}tag7$$
Where G is the Catalan's constant.
How can we prove these conjectures?
sequences-and-series
$endgroup$
I was looking at this Ramanujan phi-Function
Let:
$$R(a)=sum_{k=1}^{infty}frac{1}{(ak)^3-ak}tag1$$
and this paper of the form,
$$sum_{k=1}^{infty}frac{{2k choose k}}{4^k}tag2$$ and
$$sum_{k=1}^{infty}left[frac{{2k choose k}}{4^k}right]^2tag 3$$
By combining them together we have
$$S(a)=sum_{k=1}^{infty}frac{{2k choose k}}{4^k}cdot frac{1}{(ak)^3-ak}tag4$$ and
$$T(a)=sum_{k=1}^{infty}left[frac{{2k choose k}}{4^k}right]^2cdot frac{1}{(ak)^3-ak}tag5$$
The conjectured closed form for $(4)$ and $(5)$, where $a=2$ are
$$S(2)=sum_{k=1}^{infty}frac{{2k choose k}}{4^k}cdot frac{1}{(2k)^3-2k}=frac{pi}{4}-ln(2)tag6$$ and
$$T(2)=sum_{k=1}^{infty}left[frac{{2k choose k}}{4^k}right]^2cdot frac{1}{(2k)^3-2k}=frac{6G-piln(4)-1}{pi}tag7$$
Where G is the Catalan's constant.
How can we prove these conjectures?
sequences-and-series
sequences-and-series
edited Jan 3 at 10:10
user583851
asked Jan 3 at 9:21
user583851user583851
518110
518110
$begingroup$
Is it ${2n choose n}^2 4^{-2n}$ or ${2k choose k}^2 4^{-2k}$ in the headline?
$endgroup$
– Andreas
Jan 3 at 9:43
$begingroup$
Why don't you try the methods of the paper? ;)
$endgroup$
– Jack D'Aurizio
Jan 3 at 17:04
$begingroup$
Please notice that $(2)$ is actually a divergent series, since $frac{1}{4^n}binom{2n}{n}simfrac{1}{sqrt{pi n}}$, so "this paper of the form divergent series" does not make much sense. Better to say "this paper about the weight $frac{1}{4^n}binom{2n}{n}$ and its square".
$endgroup$
– Jack D'Aurizio
Jan 3 at 17:20
$begingroup$
$(3)$ is also a divergent series (same reason), while $$sum_{ngeq 0}left[frac{1}{4^n}binom{2n}{n}right]^3$$ is related to the lemniscate constant (hence to $Gammaleft(frac{1}{4}right)$) by many good reasons; among them, Clausen's formula for the square of particular $phantom{}_2 F_1$s.
$endgroup$
– Jack D'Aurizio
Jan 3 at 17:32
$begingroup$
$$sum_{ngeq 0}left[frac{1}{4^n}binom{2n}{n}right]^4$$ is much more obscure: see math.stackexchange.com/questions/2506266/…
$endgroup$
– Jack D'Aurizio
Jan 3 at 17:34
add a comment |
$begingroup$
Is it ${2n choose n}^2 4^{-2n}$ or ${2k choose k}^2 4^{-2k}$ in the headline?
$endgroup$
– Andreas
Jan 3 at 9:43
$begingroup$
Why don't you try the methods of the paper? ;)
$endgroup$
– Jack D'Aurizio
Jan 3 at 17:04
$begingroup$
Please notice that $(2)$ is actually a divergent series, since $frac{1}{4^n}binom{2n}{n}simfrac{1}{sqrt{pi n}}$, so "this paper of the form divergent series" does not make much sense. Better to say "this paper about the weight $frac{1}{4^n}binom{2n}{n}$ and its square".
$endgroup$
– Jack D'Aurizio
Jan 3 at 17:20
$begingroup$
$(3)$ is also a divergent series (same reason), while $$sum_{ngeq 0}left[frac{1}{4^n}binom{2n}{n}right]^3$$ is related to the lemniscate constant (hence to $Gammaleft(frac{1}{4}right)$) by many good reasons; among them, Clausen's formula for the square of particular $phantom{}_2 F_1$s.
$endgroup$
– Jack D'Aurizio
Jan 3 at 17:32
$begingroup$
$$sum_{ngeq 0}left[frac{1}{4^n}binom{2n}{n}right]^4$$ is much more obscure: see math.stackexchange.com/questions/2506266/…
$endgroup$
– Jack D'Aurizio
Jan 3 at 17:34
$begingroup$
Is it ${2n choose n}^2 4^{-2n}$ or ${2k choose k}^2 4^{-2k}$ in the headline?
$endgroup$
– Andreas
Jan 3 at 9:43
$begingroup$
Is it ${2n choose n}^2 4^{-2n}$ or ${2k choose k}^2 4^{-2k}$ in the headline?
$endgroup$
– Andreas
Jan 3 at 9:43
$begingroup$
Why don't you try the methods of the paper? ;)
$endgroup$
– Jack D'Aurizio
Jan 3 at 17:04
$begingroup$
Why don't you try the methods of the paper? ;)
$endgroup$
– Jack D'Aurizio
Jan 3 at 17:04
$begingroup$
Please notice that $(2)$ is actually a divergent series, since $frac{1}{4^n}binom{2n}{n}simfrac{1}{sqrt{pi n}}$, so "this paper of the form divergent series" does not make much sense. Better to say "this paper about the weight $frac{1}{4^n}binom{2n}{n}$ and its square".
$endgroup$
– Jack D'Aurizio
Jan 3 at 17:20
$begingroup$
Please notice that $(2)$ is actually a divergent series, since $frac{1}{4^n}binom{2n}{n}simfrac{1}{sqrt{pi n}}$, so "this paper of the form divergent series" does not make much sense. Better to say "this paper about the weight $frac{1}{4^n}binom{2n}{n}$ and its square".
$endgroup$
– Jack D'Aurizio
Jan 3 at 17:20
$begingroup$
$(3)$ is also a divergent series (same reason), while $$sum_{ngeq 0}left[frac{1}{4^n}binom{2n}{n}right]^3$$ is related to the lemniscate constant (hence to $Gammaleft(frac{1}{4}right)$) by many good reasons; among them, Clausen's formula for the square of particular $phantom{}_2 F_1$s.
$endgroup$
– Jack D'Aurizio
Jan 3 at 17:32
$begingroup$
$(3)$ is also a divergent series (same reason), while $$sum_{ngeq 0}left[frac{1}{4^n}binom{2n}{n}right]^3$$ is related to the lemniscate constant (hence to $Gammaleft(frac{1}{4}right)$) by many good reasons; among them, Clausen's formula for the square of particular $phantom{}_2 F_1$s.
$endgroup$
– Jack D'Aurizio
Jan 3 at 17:32
$begingroup$
$$sum_{ngeq 0}left[frac{1}{4^n}binom{2n}{n}right]^4$$ is much more obscure: see math.stackexchange.com/questions/2506266/…
$endgroup$
– Jack D'Aurizio
Jan 3 at 17:34
$begingroup$
$$sum_{ngeq 0}left[frac{1}{4^n}binom{2n}{n}right]^4$$ is much more obscure: see math.stackexchange.com/questions/2506266/…
$endgroup$
– Jack D'Aurizio
Jan 3 at 17:34
add a comment |
1 Answer
1
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$begingroup$
Let us tackle $(4)$ first. The ordinary generating function for $left[frac{1}{4^n}binom{2n}{n}right]$ is $frac{1}{sqrt{1-x}}$ and
$$ frac{1}{(an)^3-an}=frac{1}{(an-1)an(an+1)} = int_{0}^{1} x^{an}frac{(1-x)^2}{2x^2},dx $$
hence
$$ S(a) = int_{0}^{1}left(frac{1}{sqrt{1-x^a}}-1right)cdotfrac{(1-x)^2}{2x^2},dx $$
and similarly, since the ordinary generating function for $left[frac{1}{4^n}binom{2n}{n}right]^2$ is $frac{2}{pi}K(x)$,
$$ T(a) = frac{2}{pi}int_{0}^{1}left(K(x^a)-frac{pi}{2}right)frac{(1-x)^2}{2x^2},dx. $$
For specific values of $a$ (like $a=2$ or $a=4$) integration by parts, suitable substitutions and known FL-expansions greatly simplify the underlying hypergeometric structure of these integrals. I won't expect a simple closed form in the general case.
(6) and (7) are proved in the mentioned paper, not just conjectured.
$endgroup$
add a comment |
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1 Answer
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$begingroup$
Let us tackle $(4)$ first. The ordinary generating function for $left[frac{1}{4^n}binom{2n}{n}right]$ is $frac{1}{sqrt{1-x}}$ and
$$ frac{1}{(an)^3-an}=frac{1}{(an-1)an(an+1)} = int_{0}^{1} x^{an}frac{(1-x)^2}{2x^2},dx $$
hence
$$ S(a) = int_{0}^{1}left(frac{1}{sqrt{1-x^a}}-1right)cdotfrac{(1-x)^2}{2x^2},dx $$
and similarly, since the ordinary generating function for $left[frac{1}{4^n}binom{2n}{n}right]^2$ is $frac{2}{pi}K(x)$,
$$ T(a) = frac{2}{pi}int_{0}^{1}left(K(x^a)-frac{pi}{2}right)frac{(1-x)^2}{2x^2},dx. $$
For specific values of $a$ (like $a=2$ or $a=4$) integration by parts, suitable substitutions and known FL-expansions greatly simplify the underlying hypergeometric structure of these integrals. I won't expect a simple closed form in the general case.
(6) and (7) are proved in the mentioned paper, not just conjectured.
$endgroup$
add a comment |
$begingroup$
Let us tackle $(4)$ first. The ordinary generating function for $left[frac{1}{4^n}binom{2n}{n}right]$ is $frac{1}{sqrt{1-x}}$ and
$$ frac{1}{(an)^3-an}=frac{1}{(an-1)an(an+1)} = int_{0}^{1} x^{an}frac{(1-x)^2}{2x^2},dx $$
hence
$$ S(a) = int_{0}^{1}left(frac{1}{sqrt{1-x^a}}-1right)cdotfrac{(1-x)^2}{2x^2},dx $$
and similarly, since the ordinary generating function for $left[frac{1}{4^n}binom{2n}{n}right]^2$ is $frac{2}{pi}K(x)$,
$$ T(a) = frac{2}{pi}int_{0}^{1}left(K(x^a)-frac{pi}{2}right)frac{(1-x)^2}{2x^2},dx. $$
For specific values of $a$ (like $a=2$ or $a=4$) integration by parts, suitable substitutions and known FL-expansions greatly simplify the underlying hypergeometric structure of these integrals. I won't expect a simple closed form in the general case.
(6) and (7) are proved in the mentioned paper, not just conjectured.
$endgroup$
add a comment |
$begingroup$
Let us tackle $(4)$ first. The ordinary generating function for $left[frac{1}{4^n}binom{2n}{n}right]$ is $frac{1}{sqrt{1-x}}$ and
$$ frac{1}{(an)^3-an}=frac{1}{(an-1)an(an+1)} = int_{0}^{1} x^{an}frac{(1-x)^2}{2x^2},dx $$
hence
$$ S(a) = int_{0}^{1}left(frac{1}{sqrt{1-x^a}}-1right)cdotfrac{(1-x)^2}{2x^2},dx $$
and similarly, since the ordinary generating function for $left[frac{1}{4^n}binom{2n}{n}right]^2$ is $frac{2}{pi}K(x)$,
$$ T(a) = frac{2}{pi}int_{0}^{1}left(K(x^a)-frac{pi}{2}right)frac{(1-x)^2}{2x^2},dx. $$
For specific values of $a$ (like $a=2$ or $a=4$) integration by parts, suitable substitutions and known FL-expansions greatly simplify the underlying hypergeometric structure of these integrals. I won't expect a simple closed form in the general case.
(6) and (7) are proved in the mentioned paper, not just conjectured.
$endgroup$
Let us tackle $(4)$ first. The ordinary generating function for $left[frac{1}{4^n}binom{2n}{n}right]$ is $frac{1}{sqrt{1-x}}$ and
$$ frac{1}{(an)^3-an}=frac{1}{(an-1)an(an+1)} = int_{0}^{1} x^{an}frac{(1-x)^2}{2x^2},dx $$
hence
$$ S(a) = int_{0}^{1}left(frac{1}{sqrt{1-x^a}}-1right)cdotfrac{(1-x)^2}{2x^2},dx $$
and similarly, since the ordinary generating function for $left[frac{1}{4^n}binom{2n}{n}right]^2$ is $frac{2}{pi}K(x)$,
$$ T(a) = frac{2}{pi}int_{0}^{1}left(K(x^a)-frac{pi}{2}right)frac{(1-x)^2}{2x^2},dx. $$
For specific values of $a$ (like $a=2$ or $a=4$) integration by parts, suitable substitutions and known FL-expansions greatly simplify the underlying hypergeometric structure of these integrals. I won't expect a simple closed form in the general case.
(6) and (7) are proved in the mentioned paper, not just conjectured.
edited Jan 3 at 17:22
answered Jan 3 at 17:16
Jack D'AurizioJack D'Aurizio
292k33284672
292k33284672
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$begingroup$
Is it ${2n choose n}^2 4^{-2n}$ or ${2k choose k}^2 4^{-2k}$ in the headline?
$endgroup$
– Andreas
Jan 3 at 9:43
$begingroup$
Why don't you try the methods of the paper? ;)
$endgroup$
– Jack D'Aurizio
Jan 3 at 17:04
$begingroup$
Please notice that $(2)$ is actually a divergent series, since $frac{1}{4^n}binom{2n}{n}simfrac{1}{sqrt{pi n}}$, so "this paper of the form divergent series" does not make much sense. Better to say "this paper about the weight $frac{1}{4^n}binom{2n}{n}$ and its square".
$endgroup$
– Jack D'Aurizio
Jan 3 at 17:20
$begingroup$
$(3)$ is also a divergent series (same reason), while $$sum_{ngeq 0}left[frac{1}{4^n}binom{2n}{n}right]^3$$ is related to the lemniscate constant (hence to $Gammaleft(frac{1}{4}right)$) by many good reasons; among them, Clausen's formula for the square of particular $phantom{}_2 F_1$s.
$endgroup$
– Jack D'Aurizio
Jan 3 at 17:32
$begingroup$
$$sum_{ngeq 0}left[frac{1}{4^n}binom{2n}{n}right]^4$$ is much more obscure: see math.stackexchange.com/questions/2506266/…
$endgroup$
– Jack D'Aurizio
Jan 3 at 17:34