Generalize $sum_{n=1}^{infty}frac{H_n}{(n+1+x)(n+2+x)cdots(n+k+1+x)}$
$begingroup$
I was looking at this wolfram site on section [25]
A general identity due to B. Cloitre (pers. comm., Jan. 7, 2006)
where $kge1$
$$sum_{n=1}^{infty}frac{H_n}{(n+1)(n+2)cdots(n+k+1)}=frac{1}{k!k^2}tag1$$
I was trying to generalize $(1)$
Let generalize $(1)$,
where $xge0$
$$sum_{n=1}^{infty}frac{H_n}{(n+1+x)(n+2+x)cdots(n+k+1+x)}tag2$$
and hoping to find a closed form but I could only got a partial of it
$k=1$
$$sum_{n=1}^{infty}frac{H_n}{(n+1+x)(n+2+x)}=-frac{H_x}{x+1}-(H_x)^2+(H_{x+1})^2tag3$$
How do we go about to find the closed form for $(2)$?
I am assuming $(2)$, may take the closed form of
$$sum_{n=1}^{infty}frac{H_n}{(n+1+x)(n+2+x)cdots(n+k+1+x)}=G(x)-sum_{j=0}^{k}(-1)^j{k choose j}(H_{x+j})^2tag4$$
sequences-and-series
asked Jan 2 at 20:33
user583851user583851
508110
$endgroup$
add a comment |
$begingroup$
I was looking at this wolfram site on section [25]
A general identity due to B. Cloitre (pers. comm., Jan. 7, 2006)
where $kge1$
$$sum_{n=1}^{infty}frac{H_n}{(n+1)(n+2)cdots(n+k+1)}=frac{1}{k!k^2}tag1$$
I was trying to generalize $(1)$
Let generalize $(1)$,
where $xge0$
$$sum_{n=1}^{infty}frac{H_n}{(n+1+x)(n+2+x)cdots(n+k+1+x)}tag2$$
and hoping to find a closed form but I could only got a partial of it
$k=1$
$$sum_{n=1}^{infty}frac{H_n}{(n+1+x)(n+2+x)}=-frac{H_x}{x+1}-(H_x)^2+(H_{x+1})^2tag3$$
How do we go about to find the closed form for $(2)$?
I am assuming $(2)$, may take the closed form of
$$sum_{n=1}^{infty}frac{H_n}{(n+1+x)(n+2+x)cdots(n+k+1+x)}=G(x)-sum_{j=0}^{k}(-1)^j{k choose j}(H_{x+j})^2tag4$$
sequences-and-series
asked Jan 2 at 20:33
user583851user583851
508110
$endgroup$
add a comment |
$begingroup$
I was looking at this wolfram site on section [25]
A general identity due to B. Cloitre (pers. comm., Jan. 7, 2006)
where $kge1$
$$sum_{n=1}^{infty}frac{H_n}{(n+1)(n+2)cdots(n+k+1)}=frac{1}{k!k^2}tag1$$
I was trying to generalize $(1)$
Let generalize $(1)$,
where $xge0$
$$sum_{n=1}^{infty}frac{H_n}{(n+1+x)(n+2+x)cdots(n+k+1+x)}tag2$$
and hoping to find a closed form but I could only got a partial of it
$k=1$
$$sum_{n=1}^{infty}frac{H_n}{(n+1+x)(n+2+x)}=-frac{H_x}{x+1}-(H_x)^2+(H_{x+1})^2tag3$$
How do we go about to find the closed form for $(2)$?
I am assuming $(2)$, may take the closed form of
$$sum_{n=1}^{infty}frac{H_n}{(n+1+x)(n+2+x)cdots(n+k+1+x)}=G(x)-sum_{j=0}^{k}(-1)^j{k choose j}(H_{x+j})^2tag4$$
sequences-and-series
asked Jan 2 at 20:33
user583851user583851
508110
$endgroup$
I was looking at this wolfram site on section [25]
A general identity due to B. Cloitre (pers. comm., Jan. 7, 2006)
where $kge1$
$$sum_{n=1}^{infty}frac{H_n}{(n+1)(n+2)cdots(n+k+1)}=frac{1}{k!k^2}tag1$$
I was trying to generalize $(1)$
Let generalize $(1)$,
where $xge0$
$$sum_{n=1}^{infty}frac{H_n}{(n+1+x)(n+2+x)cdots(n+k+1+x)}tag2$$
and hoping to find a closed form but I could only got a partial of it
$k=1$
$$sum_{n=1}^{infty}frac{H_n}{(n+1+x)(n+2+x)}=-frac{H_x}{x+1}-(H_x)^2+(H_{x+1})^2tag3$$
How do we go about to find the closed form for $(2)$?
I am assuming $(2)$, may take the closed form of
$$sum_{n=1}^{infty}frac{H_n}{(n+1+x)(n+2+x)cdots(n+k+1+x)}=G(x)-sum_{j=0}^{k}(-1)^j{k choose j}(H_{x+j})^2tag4$$
sequences-and-series
sequences-and-series
asked Jan 2 at 20:33
user583851user583851
508110
asked Jan 2 at 20:33
user583851user583851
508110
edited Jan 2 at 20:53
user583851
asked Jan 2 at 20:33
user583851user583851
508110
asked Jan 2 at 20:33
user583851user583851
508110
asked Jan 2 at 20:33
user583851user583851
508110
508110
add a comment |
add a comment |
1 Answer
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By summation by parts the computation of $(2)$ boils down to the computation of
$$ sum_{ngeq 1}frac{1}{(n+1)(n+1+x)(n+2+x)cdots(n+k+x)}=sum_{ngeq 1}frac{Gamma(n+1+x)}{(n+1)Gamma(n+k+x)} $$
or
$$ frac{1}{Gamma(k-1)}sum_{ngeq 1}frac{B(n+1+x,k-1)}{n+1}=frac{1}{Gamma(k-1)}int_{0}^{1}sum_{ngeq 1}frac{(1-z)^{k-2}z^{n+x+2}}{n+1},dz $$
or
$$ frac{1}{Gamma(k-1)}int_{0}^{1}(1-z)^{k-2}z^{1+x}(-z-log(1-z)),dz $$
which only depends on the Beta function and its partial derivatives. In particular the last expression equals
$$ -frac{Gamma(x+2)Gamma(k-1)}{Gamma(x+k+2)}left[2+x+(x+k+1)(H_{k-2}-H_{k+x})right]. $$
answered Jan 2 at 21:54
Jack D'AurizioJack D'Aurizio
291k33284666
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1 Answer
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1 Answer
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$begingroup$
By summation by parts the computation of $(2)$ boils down to the computation of
$$ sum_{ngeq 1}frac{1}{(n+1)(n+1+x)(n+2+x)cdots(n+k+x)}=sum_{ngeq 1}frac{Gamma(n+1+x)}{(n+1)Gamma(n+k+x)} $$
or
$$ frac{1}{Gamma(k-1)}sum_{ngeq 1}frac{B(n+1+x,k-1)}{n+1}=frac{1}{Gamma(k-1)}int_{0}^{1}sum_{ngeq 1}frac{(1-z)^{k-2}z^{n+x+2}}{n+1},dz $$
or
$$ frac{1}{Gamma(k-1)}int_{0}^{1}(1-z)^{k-2}z^{1+x}(-z-log(1-z)),dz $$
which only depends on the Beta function and its partial derivatives. In particular the last expression equals
$$ -frac{Gamma(x+2)Gamma(k-1)}{Gamma(x+k+2)}left[2+x+(x+k+1)(H_{k-2}-H_{k+x})right]. $$
answered Jan 2 at 21:54
Jack D'AurizioJack D'Aurizio
291k33284666
$endgroup$
add a comment |
$begingroup$
By summation by parts the computation of $(2)$ boils down to the computation of
$$ sum_{ngeq 1}frac{1}{(n+1)(n+1+x)(n+2+x)cdots(n+k+x)}=sum_{ngeq 1}frac{Gamma(n+1+x)}{(n+1)Gamma(n+k+x)} $$
or
$$ frac{1}{Gamma(k-1)}sum_{ngeq 1}frac{B(n+1+x,k-1)}{n+1}=frac{1}{Gamma(k-1)}int_{0}^{1}sum_{ngeq 1}frac{(1-z)^{k-2}z^{n+x+2}}{n+1},dz $$
or
$$ frac{1}{Gamma(k-1)}int_{0}^{1}(1-z)^{k-2}z^{1+x}(-z-log(1-z)),dz $$
which only depends on the Beta function and its partial derivatives. In particular the last expression equals
$$ -frac{Gamma(x+2)Gamma(k-1)}{Gamma(x+k+2)}left[2+x+(x+k+1)(H_{k-2}-H_{k+x})right]. $$
answered Jan 2 at 21:54
Jack D'AurizioJack D'Aurizio
291k33284666
$endgroup$
add a comment |
$begingroup$
By summation by parts the computation of $(2)$ boils down to the computation of
$$ sum_{ngeq 1}frac{1}{(n+1)(n+1+x)(n+2+x)cdots(n+k+x)}=sum_{ngeq 1}frac{Gamma(n+1+x)}{(n+1)Gamma(n+k+x)} $$
or
$$ frac{1}{Gamma(k-1)}sum_{ngeq 1}frac{B(n+1+x,k-1)}{n+1}=frac{1}{Gamma(k-1)}int_{0}^{1}sum_{ngeq 1}frac{(1-z)^{k-2}z^{n+x+2}}{n+1},dz $$
or
$$ frac{1}{Gamma(k-1)}int_{0}^{1}(1-z)^{k-2}z^{1+x}(-z-log(1-z)),dz $$
which only depends on the Beta function and its partial derivatives. In particular the last expression equals
$$ -frac{Gamma(x+2)Gamma(k-1)}{Gamma(x+k+2)}left[2+x+(x+k+1)(H_{k-2}-H_{k+x})right]. $$
answered Jan 2 at 21:54
Jack D'AurizioJack D'Aurizio
291k33284666
$endgroup$
By summation by parts the computation of $(2)$ boils down to the computation of
$$ sum_{ngeq 1}frac{1}{(n+1)(n+1+x)(n+2+x)cdots(n+k+x)}=sum_{ngeq 1}frac{Gamma(n+1+x)}{(n+1)Gamma(n+k+x)} $$
or
$$ frac{1}{Gamma(k-1)}sum_{ngeq 1}frac{B(n+1+x,k-1)}{n+1}=frac{1}{Gamma(k-1)}int_{0}^{1}sum_{ngeq 1}frac{(1-z)^{k-2}z^{n+x+2}}{n+1},dz $$
or
$$ frac{1}{Gamma(k-1)}int_{0}^{1}(1-z)^{k-2}z^{1+x}(-z-log(1-z)),dz $$
which only depends on the Beta function and its partial derivatives. In particular the last expression equals
$$ -frac{Gamma(x+2)Gamma(k-1)}{Gamma(x+k+2)}left[2+x+(x+k+1)(H_{k-2}-H_{k+x})right]. $$
answered Jan 2 at 21:54
Jack D'AurizioJack D'Aurizio
291k33284666
answered Jan 2 at 21:54
Jack D'AurizioJack D'Aurizio
291k33284666
answered Jan 2 at 21:54
Jack D'AurizioJack D'Aurizio
291k33284666
answered Jan 2 at 21:54
Jack D'AurizioJack D'Aurizio
291k33284666
291k33284666
add a comment |
add a comment |
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