Generalize $sum_{n=1}^{infty}frac{H_n^2-H_n^{(2)}}{(n+1)(n+2)}=2$
$begingroup$
In this paper on section [5],
Recently J. Choi [4, Corollary 3] proved a sequence of identities:
$$sum_{n=1}^{infty}frac{H_n^2-H_n^{(2)}}{(n+1)(n+2)}=2tag1$$
Let just generalize $(1)$
$$sum_{n=1}^{infty}frac{H_n^2-H_n^{(2)}}{(n+1)(n+2)cdots(n+k)}tag2$$
where $kge 2$
We conjectured the closed form of $(2)$ to be
$$sum_{n=1}^{infty}frac{H_n^2-H_n^{(2)}}{(n+1)(n+2)cdots(n+k)}=frac{2^k}{(2k-2)!!}cdot frac{1}{(k-1)^3}=frac{2}{(k-1)^3(k-1)!}tag3$$
How may we prove $(3)$?
sequences-and-series harmonic-functions
asked Jan 13 at 8:25
user583851user583851
518110
$endgroup$
add a comment |
$begingroup$
In this paper on section [5],
Recently J. Choi [4, Corollary 3] proved a sequence of identities:
$$sum_{n=1}^{infty}frac{H_n^2-H_n^{(2)}}{(n+1)(n+2)}=2tag1$$
Let just generalize $(1)$
$$sum_{n=1}^{infty}frac{H_n^2-H_n^{(2)}}{(n+1)(n+2)cdots(n+k)}tag2$$
where $kge 2$
We conjectured the closed form of $(2)$ to be
$$sum_{n=1}^{infty}frac{H_n^2-H_n^{(2)}}{(n+1)(n+2)cdots(n+k)}=frac{2^k}{(2k-2)!!}cdot frac{1}{(k-1)^3}=frac{2}{(k-1)^3(k-1)!}tag3$$
How may we prove $(3)$?
sequences-and-series harmonic-functions
asked Jan 13 at 8:25
user583851user583851
518110
$endgroup$
add a comment |
$begingroup$
In this paper on section [5],
Recently J. Choi [4, Corollary 3] proved a sequence of identities:
$$sum_{n=1}^{infty}frac{H_n^2-H_n^{(2)}}{(n+1)(n+2)}=2tag1$$
Let just generalize $(1)$
$$sum_{n=1}^{infty}frac{H_n^2-H_n^{(2)}}{(n+1)(n+2)cdots(n+k)}tag2$$
where $kge 2$
We conjectured the closed form of $(2)$ to be
$$sum_{n=1}^{infty}frac{H_n^2-H_n^{(2)}}{(n+1)(n+2)cdots(n+k)}=frac{2^k}{(2k-2)!!}cdot frac{1}{(k-1)^3}=frac{2}{(k-1)^3(k-1)!}tag3$$
How may we prove $(3)$?
sequences-and-series harmonic-functions
asked Jan 13 at 8:25
user583851user583851
518110
$endgroup$
In this paper on section [5],
Recently J. Choi [4, Corollary 3] proved a sequence of identities:
$$sum_{n=1}^{infty}frac{H_n^2-H_n^{(2)}}{(n+1)(n+2)}=2tag1$$
Let just generalize $(1)$
$$sum_{n=1}^{infty}frac{H_n^2-H_n^{(2)}}{(n+1)(n+2)cdots(n+k)}tag2$$
where $kge 2$
We conjectured the closed form of $(2)$ to be
$$sum_{n=1}^{infty}frac{H_n^2-H_n^{(2)}}{(n+1)(n+2)cdots(n+k)}=frac{2^k}{(2k-2)!!}cdot frac{1}{(k-1)^3}=frac{2}{(k-1)^3(k-1)!}tag3$$
How may we prove $(3)$?
sequences-and-series harmonic-functions
sequences-and-series harmonic-functions
asked Jan 13 at 8:25
user583851user583851
518110
asked Jan 13 at 8:25
user583851user583851
518110
edited Jan 13 at 20:16
user583851
asked Jan 13 at 8:25
user583851user583851
518110
asked Jan 13 at 8:25
user583851user583851
518110
asked Jan 13 at 8:25
user583851user583851
518110
518110
add a comment |
add a comment |
1 Answer
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Just realize you are computing
$$int_{0}^{1}(1-x)^n log^2(1-x),dx,qquad int_{0}^{1}(1-x)^mtext{Li}_2(x),dx $$
which are elementary integrals.
answered Jan 13 at 22:21
Jack D'AurizioJack D'Aurizio
291k33284668
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1 Answer
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$begingroup$
Just realize you are computing
$$int_{0}^{1}(1-x)^n log^2(1-x),dx,qquad int_{0}^{1}(1-x)^mtext{Li}_2(x),dx $$
which are elementary integrals.
answered Jan 13 at 22:21
Jack D'AurizioJack D'Aurizio
291k33284668
$endgroup$
add a comment |
$begingroup$
Just realize you are computing
$$int_{0}^{1}(1-x)^n log^2(1-x),dx,qquad int_{0}^{1}(1-x)^mtext{Li}_2(x),dx $$
which are elementary integrals.
answered Jan 13 at 22:21
Jack D'AurizioJack D'Aurizio
291k33284668
$endgroup$
add a comment |
$begingroup$
Just realize you are computing
$$int_{0}^{1}(1-x)^n log^2(1-x),dx,qquad int_{0}^{1}(1-x)^mtext{Li}_2(x),dx $$
which are elementary integrals.
answered Jan 13 at 22:21
Jack D'AurizioJack D'Aurizio
291k33284668
$endgroup$
Just realize you are computing
$$int_{0}^{1}(1-x)^n log^2(1-x),dx,qquad int_{0}^{1}(1-x)^mtext{Li}_2(x),dx $$
which are elementary integrals.
answered Jan 13 at 22:21
Jack D'AurizioJack D'Aurizio
291k33284668
answered Jan 13 at 22:21
Jack D'AurizioJack D'Aurizio
291k33284668
answered Jan 13 at 22:21
Jack D'AurizioJack D'Aurizio
291k33284668
answered Jan 13 at 22:21
Jack D'AurizioJack D'Aurizio
291k33284668
291k33284668
add a comment |
add a comment |
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