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Searching for a proof for a series identity
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The below identity I have found experimentally.
Question. Is this true? If so, may you provide a "slick" (or any) proof.
$$6sum_k=1^inftyfrack^2q^k(1-q^k)^2+12left(sum_k=1^inftyfrackq^k1-q^kright)^2=sum_k=1^inftyfrac(5k^3+k)q^k1-q^k.$$
nt.number-theory real-analysis sequences-and-series power-series q-identities
edited 1 hour ago
Robert Israel
40.5k46114
asked 4 hours ago
T. Amdeberhan
15.7k225119
add a comment |Â
up vote
5
down vote
favorite
The below identity I have found experimentally.
Question. Is this true? If so, may you provide a "slick" (or any) proof.
$$6sum_k=1^inftyfrack^2q^k(1-q^k)^2+12left(sum_k=1^inftyfrackq^k1-q^kright)^2=sum_k=1^inftyfrac(5k^3+k)q^k1-q^k.$$
nt.number-theory real-analysis sequences-and-series power-series q-identities
edited 1 hour ago
Robert Israel
40.5k46114
asked 4 hours ago
T. Amdeberhan
15.7k225119
add a comment |Â
up vote
5
down vote
favorite
up vote
5
down vote
favorite
The below identity I have found experimentally.
Question. Is this true? If so, may you provide a "slick" (or any) proof.
$$6sum_k=1^inftyfrack^2q^k(1-q^k)^2+12left(sum_k=1^inftyfrackq^k1-q^kright)^2=sum_k=1^inftyfrac(5k^3+k)q^k1-q^k.$$
nt.number-theory real-analysis sequences-and-series power-series q-identities
edited 1 hour ago
Robert Israel
40.5k46114
asked 4 hours ago
T. Amdeberhan
15.7k225119
The below identity I have found experimentally.
Question. Is this true? If so, may you provide a "slick" (or any) proof.
$$6sum_k=1^inftyfrack^2q^k(1-q^k)^2+12left(sum_k=1^inftyfrackq^k1-q^kright)^2=sum_k=1^inftyfrac(5k^3+k)q^k1-q^k.$$
nt.number-theory real-analysis sequences-and-series power-series q-identities
nt.number-theory real-analysis sequences-and-series power-series q-identities
edited 1 hour ago
Robert Israel
40.5k46114
asked 4 hours ago
T. Amdeberhan
15.7k225119
edited 1 hour ago
Robert Israel
40.5k46114
asked 4 hours ago
T. Amdeberhan
15.7k225119
edited 1 hour ago
Robert Israel
40.5k46114
edited 1 hour ago
Robert Israel
40.5k46114
edited 1 hour ago
Robert Israel
40.5k46114
40.5k46114
asked 4 hours ago
T. Amdeberhan
15.7k225119
asked 4 hours ago
T. Amdeberhan
15.7k225119
asked 4 hours ago
T. Amdeberhan
15.7k225119
15.7k225119
add a comment |Â
add a comment |Â
2 Answers
2
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Follow the comments of Lucia and note that
$$sum_nge 1fracn^2q^n(1-q^n)^2=qfrac,d,dqsum_nge 1fracnq^n1-q^n.$$
I believe the identity actually is the well known
$$qfrac,d,dqL=fracL^2-M12,$$
where
$$L=1-24sum_nge 1fracnq^n1-q^n;mbox
and;
M=1+240sum_nge 1fracn^3q^n1-q^n.$$
You can find it in here [https://en.wikipedia.org/wiki/Eisenstein_series].
answered 21 mins ago
Zhou
39825
add a comment |Â
up vote
1
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$$ sum_k=1^infty frack^2 q^k(1-q^k)^2 = sum_n=1^infty sigma(n) n q^n$$
$$ sum_k=1^infty frack q^k1-q^k = sum_n=1^infty sigma(n) q^n$$
$$ left(sum_k=1^infty frack q^k1-q^kright)^2 = sum_n=1^infty sum_m=1^n-1 sigma(m) sigma(n-m) q^n $$
$$ sum_k=1^infty frack^3 q^k1-q^k = sum_n=1^infty sigma_3(n) q^n $$
so your identity is saying
$$ 6 n sigma(n) + 12 sum_m=1^n-1 sigma(m)sigma(n-m) = 5 sigma_3(n) + sigma(n)$$
Hmm, surely that's got to be known. Adding number-theory to the tags.
answered 1 hour ago
Robert Israel
40.5k46114
5
Such identities are obtained by multiplying Eisenstein series and decomposing the product in terms of modular forms. See Ramanujan's paper On certain arithmetical functions which includes this identity, Table IV line 1. ramanujan.sirinudi.org/Volumes/published/ram18.pdf
– Lucia
51 mins ago
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
Follow the comments of Lucia and note that
$$sum_nge 1fracn^2q^n(1-q^n)^2=qfrac,d,dqsum_nge 1fracnq^n1-q^n.$$
I believe the identity actually is the well known
$$qfrac,d,dqL=fracL^2-M12,$$
where
$$L=1-24sum_nge 1fracnq^n1-q^n;mbox
and;
M=1+240sum_nge 1fracn^3q^n1-q^n.$$
You can find it in here [https://en.wikipedia.org/wiki/Eisenstein_series].
answered 21 mins ago
Zhou
39825
add a comment |Â
up vote
3
down vote
Follow the comments of Lucia and note that
$$sum_nge 1fracn^2q^n(1-q^n)^2=qfrac,d,dqsum_nge 1fracnq^n1-q^n.$$
I believe the identity actually is the well known
$$qfrac,d,dqL=fracL^2-M12,$$
where
$$L=1-24sum_nge 1fracnq^n1-q^n;mbox
and;
M=1+240sum_nge 1fracn^3q^n1-q^n.$$
You can find it in here [https://en.wikipedia.org/wiki/Eisenstein_series].
answered 21 mins ago
Zhou
39825
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Follow the comments of Lucia and note that
$$sum_nge 1fracn^2q^n(1-q^n)^2=qfrac,d,dqsum_nge 1fracnq^n1-q^n.$$
I believe the identity actually is the well known
$$qfrac,d,dqL=fracL^2-M12,$$
where
$$L=1-24sum_nge 1fracnq^n1-q^n;mbox
and;
M=1+240sum_nge 1fracn^3q^n1-q^n.$$
You can find it in here [https://en.wikipedia.org/wiki/Eisenstein_series].
answered 21 mins ago
Zhou
39825
Follow the comments of Lucia and note that
$$sum_nge 1fracn^2q^n(1-q^n)^2=qfrac,d,dqsum_nge 1fracnq^n1-q^n.$$
I believe the identity actually is the well known
$$qfrac,d,dqL=fracL^2-M12,$$
where
$$L=1-24sum_nge 1fracnq^n1-q^n;mbox
and;
M=1+240sum_nge 1fracn^3q^n1-q^n.$$
You can find it in here [https://en.wikipedia.org/wiki/Eisenstein_series].
answered 21 mins ago
Zhou
39825
answered 21 mins ago
Zhou
39825
answered 21 mins ago
Zhou
39825
answered 21 mins ago
Zhou
39825
39825
add a comment |Â
add a comment |Â
up vote
1
down vote
$$ sum_k=1^infty frack^2 q^k(1-q^k)^2 = sum_n=1^infty sigma(n) n q^n$$
$$ sum_k=1^infty frack q^k1-q^k = sum_n=1^infty sigma(n) q^n$$
$$ left(sum_k=1^infty frack q^k1-q^kright)^2 = sum_n=1^infty sum_m=1^n-1 sigma(m) sigma(n-m) q^n $$
$$ sum_k=1^infty frack^3 q^k1-q^k = sum_n=1^infty sigma_3(n) q^n $$
so your identity is saying
$$ 6 n sigma(n) + 12 sum_m=1^n-1 sigma(m)sigma(n-m) = 5 sigma_3(n) + sigma(n)$$
Hmm, surely that's got to be known. Adding number-theory to the tags.
answered 1 hour ago
Robert Israel
40.5k46114
5
Such identities are obtained by multiplying Eisenstein series and decomposing the product in terms of modular forms. See Ramanujan's paper On certain arithmetical functions which includes this identity, Table IV line 1. ramanujan.sirinudi.org/Volumes/published/ram18.pdf
– Lucia
51 mins ago
add a comment |Â
up vote
1
down vote
$$ sum_k=1^infty frack^2 q^k(1-q^k)^2 = sum_n=1^infty sigma(n) n q^n$$
$$ sum_k=1^infty frack q^k1-q^k = sum_n=1^infty sigma(n) q^n$$
$$ left(sum_k=1^infty frack q^k1-q^kright)^2 = sum_n=1^infty sum_m=1^n-1 sigma(m) sigma(n-m) q^n $$
$$ sum_k=1^infty frack^3 q^k1-q^k = sum_n=1^infty sigma_3(n) q^n $$
so your identity is saying
$$ 6 n sigma(n) + 12 sum_m=1^n-1 sigma(m)sigma(n-m) = 5 sigma_3(n) + sigma(n)$$
Hmm, surely that's got to be known. Adding number-theory to the tags.
answered 1 hour ago
Robert Israel
40.5k46114
5
Such identities are obtained by multiplying Eisenstein series and decomposing the product in terms of modular forms. See Ramanujan's paper On certain arithmetical functions which includes this identity, Table IV line 1. ramanujan.sirinudi.org/Volumes/published/ram18.pdf
– Lucia
51 mins ago
add a comment |Â
up vote
1
down vote
up vote
1
down vote
$$ sum_k=1^infty frack^2 q^k(1-q^k)^2 = sum_n=1^infty sigma(n) n q^n$$
$$ sum_k=1^infty frack q^k1-q^k = sum_n=1^infty sigma(n) q^n$$
$$ left(sum_k=1^infty frack q^k1-q^kright)^2 = sum_n=1^infty sum_m=1^n-1 sigma(m) sigma(n-m) q^n $$
$$ sum_k=1^infty frack^3 q^k1-q^k = sum_n=1^infty sigma_3(n) q^n $$
so your identity is saying
$$ 6 n sigma(n) + 12 sum_m=1^n-1 sigma(m)sigma(n-m) = 5 sigma_3(n) + sigma(n)$$
Hmm, surely that's got to be known. Adding number-theory to the tags.
answered 1 hour ago
Robert Israel
40.5k46114
$$ sum_k=1^infty frack^2 q^k(1-q^k)^2 = sum_n=1^infty sigma(n) n q^n$$
$$ sum_k=1^infty frack q^k1-q^k = sum_n=1^infty sigma(n) q^n$$
$$ left(sum_k=1^infty frack q^k1-q^kright)^2 = sum_n=1^infty sum_m=1^n-1 sigma(m) sigma(n-m) q^n $$
$$ sum_k=1^infty frack^3 q^k1-q^k = sum_n=1^infty sigma_3(n) q^n $$
so your identity is saying
$$ 6 n sigma(n) + 12 sum_m=1^n-1 sigma(m)sigma(n-m) = 5 sigma_3(n) + sigma(n)$$
Hmm, surely that's got to be known. Adding number-theory to the tags.
answered 1 hour ago
Robert Israel
40.5k46114
answered 1 hour ago
Robert Israel
40.5k46114
answered 1 hour ago
Robert Israel
40.5k46114
answered 1 hour ago
Robert Israel
40.5k46114
40.5k46114
5
Such identities are obtained by multiplying Eisenstein series and decomposing the product in terms of modular forms. See Ramanujan's paper On certain arithmetical functions which includes this identity, Table IV line 1. ramanujan.sirinudi.org/Volumes/published/ram18.pdf
– Lucia
51 mins ago
add a comment |Â
5
Such identities are obtained by multiplying Eisenstein series and decomposing the product in terms of modular forms. See Ramanujan's paper On certain arithmetical functions which includes this identity, Table IV line 1. ramanujan.sirinudi.org/Volumes/published/ram18.pdf
– Lucia
51 mins ago
5
5
Such identities are obtained by multiplying Eisenstein series and decomposing the product in terms of modular forms. See Ramanujan's paper On certain arithmetical functions which includes this identity, Table IV line 1. ramanujan.sirinudi.org/Volumes/published/ram18.pdf
– Lucia
51 mins ago
Such identities are obtained by multiplying Eisenstein series and decomposing the product in terms of modular forms. See Ramanujan's paper On certain arithmetical functions which includes this identity, Table IV line 1. ramanujan.sirinudi.org/Volumes/published/ram18.pdf
– Lucia
51 mins ago
add a comment |Â
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