Do harmonic numbers have a “closed-form” expression?
One of the joys of high-school mathematics is summing a complicated series to get a “closed-form” expression. And of course many of us have tried summing the harmonic series $H_n =sum limits_{k leq n} frac{1}{k}$, and failed. But should we necessarily fail?
More precisely, is it known that $H_n$ cannot be written in terms of the elementary functions, say, the rational functions, $exp(x)$ and $ln x$? If so, how is such a theorem proved?
Note. When I started writing the question, I was going to ask if it is known that the harmonic function cannot be represented simply as a rational function? But this is easy to see, since $H_n$ grows like $ln n+O(1)$, whereas no rational function grows logarithmically.
Added note: This earlier question asks a similar question for “elementary integration”. I guess I am asking if there is an analogous theory of “elementary summation”.
abstract-algebra number-theory functions closed-form harmonic-numbers
|
show 2 more comments
One of the joys of high-school mathematics is summing a complicated series to get a “closed-form” expression. And of course many of us have tried summing the harmonic series $H_n =sum limits_{k leq n} frac{1}{k}$, and failed. But should we necessarily fail?
More precisely, is it known that $H_n$ cannot be written in terms of the elementary functions, say, the rational functions, $exp(x)$ and $ln x$? If so, how is such a theorem proved?
Note. When I started writing the question, I was going to ask if it is known that the harmonic function cannot be represented simply as a rational function? But this is easy to see, since $H_n$ grows like $ln n+O(1)$, whereas no rational function grows logarithmically.
Added note: This earlier question asks a similar question for “elementary integration”. I guess I am asking if there is an analogous theory of “elementary summation”.
abstract-algebra number-theory functions closed-form harmonic-numbers
1
So what is the question?
– Ross Millikan
Jul 20 '11 at 5:07
8
Ross, the second paragraph explicitly and directly states the question. If the first paragraph is too verbose, I can trim it a bit.
– Srivatsan
Jul 20 '11 at 5:25
6
There is an expression which might loosely be called "closed form": $H_n = Psi(n+1) + gamma$, where $Psi$ is the "digamma" function $Psi(x) = frac{d}{dx} ln Gamma(x)$. I don't know how to prove that $Psi$, or $Gamma$ for that matter, is not elementary.
– Robert Israel
Jul 20 '11 at 5:58
In addition to $Psi$, I am not sure if using everyone will agree with using constant $gamma$ which itself doesn't have a nice form. But I am ok with it. (After all, I did allow $e$!)
– Srivatsan
Jul 20 '11 at 6:12
4
FWIW, I consider harmonic numbers as closed forms in themselves, just as I consider $n!$ to be the closed form for $prod_{k=1}^n k$ and $(a)_n$ to be the closed form of $prod_{k=0}^{n-1} (a+k)$...
– J. M. is not a mathematician
Jul 20 '11 at 11:19
|
show 2 more comments
One of the joys of high-school mathematics is summing a complicated series to get a “closed-form” expression. And of course many of us have tried summing the harmonic series $H_n =sum limits_{k leq n} frac{1}{k}$, and failed. But should we necessarily fail?
More precisely, is it known that $H_n$ cannot be written in terms of the elementary functions, say, the rational functions, $exp(x)$ and $ln x$? If so, how is such a theorem proved?
Note. When I started writing the question, I was going to ask if it is known that the harmonic function cannot be represented simply as a rational function? But this is easy to see, since $H_n$ grows like $ln n+O(1)$, whereas no rational function grows logarithmically.
Added note: This earlier question asks a similar question for “elementary integration”. I guess I am asking if there is an analogous theory of “elementary summation”.
abstract-algebra number-theory functions closed-form harmonic-numbers
One of the joys of high-school mathematics is summing a complicated series to get a “closed-form” expression. And of course many of us have tried summing the harmonic series $H_n =sum limits_{k leq n} frac{1}{k}$, and failed. But should we necessarily fail?
More precisely, is it known that $H_n$ cannot be written in terms of the elementary functions, say, the rational functions, $exp(x)$ and $ln x$? If so, how is such a theorem proved?
Note. When I started writing the question, I was going to ask if it is known that the harmonic function cannot be represented simply as a rational function? But this is easy to see, since $H_n$ grows like $ln n+O(1)$, whereas no rational function grows logarithmically.
Added note: This earlier question asks a similar question for “elementary integration”. I guess I am asking if there is an analogous theory of “elementary summation”.
abstract-algebra number-theory functions closed-form harmonic-numbers
abstract-algebra number-theory functions closed-form harmonic-numbers
edited Apr 13 '17 at 12:20
Community♦
1
1
asked Jul 20 '11 at 5:03
SrivatsanSrivatsan
20.9k371125
20.9k371125
1
So what is the question?
– Ross Millikan
Jul 20 '11 at 5:07
8
Ross, the second paragraph explicitly and directly states the question. If the first paragraph is too verbose, I can trim it a bit.
– Srivatsan
Jul 20 '11 at 5:25
6
There is an expression which might loosely be called "closed form": $H_n = Psi(n+1) + gamma$, where $Psi$ is the "digamma" function $Psi(x) = frac{d}{dx} ln Gamma(x)$. I don't know how to prove that $Psi$, or $Gamma$ for that matter, is not elementary.
– Robert Israel
Jul 20 '11 at 5:58
In addition to $Psi$, I am not sure if using everyone will agree with using constant $gamma$ which itself doesn't have a nice form. But I am ok with it. (After all, I did allow $e$!)
– Srivatsan
Jul 20 '11 at 6:12
4
FWIW, I consider harmonic numbers as closed forms in themselves, just as I consider $n!$ to be the closed form for $prod_{k=1}^n k$ and $(a)_n$ to be the closed form of $prod_{k=0}^{n-1} (a+k)$...
– J. M. is not a mathematician
Jul 20 '11 at 11:19
|
show 2 more comments
1
So what is the question?
– Ross Millikan
Jul 20 '11 at 5:07
8
Ross, the second paragraph explicitly and directly states the question. If the first paragraph is too verbose, I can trim it a bit.
– Srivatsan
Jul 20 '11 at 5:25
6
There is an expression which might loosely be called "closed form": $H_n = Psi(n+1) + gamma$, where $Psi$ is the "digamma" function $Psi(x) = frac{d}{dx} ln Gamma(x)$. I don't know how to prove that $Psi$, or $Gamma$ for that matter, is not elementary.
– Robert Israel
Jul 20 '11 at 5:58
In addition to $Psi$, I am not sure if using everyone will agree with using constant $gamma$ which itself doesn't have a nice form. But I am ok with it. (After all, I did allow $e$!)
– Srivatsan
Jul 20 '11 at 6:12
4
FWIW, I consider harmonic numbers as closed forms in themselves, just as I consider $n!$ to be the closed form for $prod_{k=1}^n k$ and $(a)_n$ to be the closed form of $prod_{k=0}^{n-1} (a+k)$...
– J. M. is not a mathematician
Jul 20 '11 at 11:19
1
1
So what is the question?
– Ross Millikan
Jul 20 '11 at 5:07
So what is the question?
– Ross Millikan
Jul 20 '11 at 5:07
8
8
Ross, the second paragraph explicitly and directly states the question. If the first paragraph is too verbose, I can trim it a bit.
– Srivatsan
Jul 20 '11 at 5:25
Ross, the second paragraph explicitly and directly states the question. If the first paragraph is too verbose, I can trim it a bit.
– Srivatsan
Jul 20 '11 at 5:25
6
6
There is an expression which might loosely be called "closed form": $H_n = Psi(n+1) + gamma$, where $Psi$ is the "digamma" function $Psi(x) = frac{d}{dx} ln Gamma(x)$. I don't know how to prove that $Psi$, or $Gamma$ for that matter, is not elementary.
– Robert Israel
Jul 20 '11 at 5:58
There is an expression which might loosely be called "closed form": $H_n = Psi(n+1) + gamma$, where $Psi$ is the "digamma" function $Psi(x) = frac{d}{dx} ln Gamma(x)$. I don't know how to prove that $Psi$, or $Gamma$ for that matter, is not elementary.
– Robert Israel
Jul 20 '11 at 5:58
In addition to $Psi$, I am not sure if using everyone will agree with using constant $gamma$ which itself doesn't have a nice form. But I am ok with it. (After all, I did allow $e$!)
– Srivatsan
Jul 20 '11 at 6:12
In addition to $Psi$, I am not sure if using everyone will agree with using constant $gamma$ which itself doesn't have a nice form. But I am ok with it. (After all, I did allow $e$!)
– Srivatsan
Jul 20 '11 at 6:12
4
4
FWIW, I consider harmonic numbers as closed forms in themselves, just as I consider $n!$ to be the closed form for $prod_{k=1}^n k$ and $(a)_n$ to be the closed form of $prod_{k=0}^{n-1} (a+k)$...
– J. M. is not a mathematician
Jul 20 '11 at 11:19
FWIW, I consider harmonic numbers as closed forms in themselves, just as I consider $n!$ to be the closed form for $prod_{k=1}^n k$ and $(a)_n$ to be the closed form of $prod_{k=0}^{n-1} (a+k)$...
– J. M. is not a mathematician
Jul 20 '11 at 11:19
|
show 2 more comments
5 Answers
5
active
oldest
votes
There is a theory of elementary summation; the phrase generally used is "summation in finite terms." An important reference is Michael Karr, Summation in finite terms, Journal of the Association for Computing Machinery 28 (1981) 305-350, DOI: 10.1145/322248.322255. Quoting,
This paper describes techniques which greatly broaden the scope of what is meant by 'finite terms'...these methods will show that the following sums have no formula as a rational function of $n$:
$$sum_{i=1}^n{1over i},quad sum_{i=1}^n{1over i^2},quad sum_{i=1}^n{2^iover i},quad sum_{i=1}^ni!$$
Undoubtedly the particular problem of $H_n$ goes back well before 1981. The references in Karr's paper may be of some help here.
9
For lazy people like me...
– J. M. is not a mathematician
Jul 20 '11 at 11:20
1
There is also this paper: Michael Karr, "Theory of summation in finite terms", Journal of Symbolic Computation 1 (1985), no. 3, 303–315. MR0849038 (89a:12016)
– lhf
Jul 21 '11 at 22:10
add a comment |
Harmonic numbers can be represented in terms of integrals of elementary functions:
$$H_n=frac{int_0^{infty} x^n e^{-x} log x ; dx}{int_0^{infty} x^n e^{-x} dx}-int_0^{infty} e^{-x} log x ; dx.$$
This formula could also be used to generalize harmonic numbers to fractional or even complex arguments. These generalized harmonic numbers retain some of their useful properties, for example,
$$H_z=H_{z-1}+frac{1}{z}.$$
9
The Harmonic numbers can also be represented by the integral $H_n = int_0^1 frac{1-x^n}{1-x} dx$
– Rob Bland
Nov 12 '15 at 2:17
add a comment |
This is probably not what you were looking for (since it isn't in terms of rational or elementary functions), but for the harmonic numbers we have
$$H_n=frac{1}{n!}left[{n+1 atop 2}right]$$
where $left[{n atop k}right]$ are the (unsigned) Stirling numbers of the first kind (page 261 from the book Concrete Mathematics by Graham, Knuth and Patashnik - second edition).
For the generalized harmonic numbers I like this formula - even though it does involve an integral and Riemann zeta...
Maybe you prefer this
add a comment |
$$H_n = frac{binom{(n+1)!+n}{n}-1}{(n+1)!}-(n+1)Biggllfloor frac{binom{(n+1)!+n}{n}-1}{(n+1)(n+1)!}Biggrrfloor $$
That's a remarkable closed form, do you have a source for a proof or more detailed elaboration on how this is derived?
– RocketNuts
May 16 '18 at 23:20
add a comment |
The following series shows the relationship between the harmonic numbers and the logarithm of odd integers.
$$
log(2n+1)=H_n+sum_{k=1}^{infty}left(sum_{i=-n}^{n}frac{1}{(2n+1)k+i}-frac{1}{k}right)
$$
https://math.stackexchange.com/a/1602945/134791
Equivalently,
$$
H_n=log(2n+1)-sum_{k=1}^{infty}left(sum_{i=-n}^{n}frac{1}{(2n+1)k+i}-frac{1}{k}right)
$$
An integral form is given by
$$H_n=log(2n+1)-int_{0}^{1} frac{x^n(1-x)}{sum_{k=0}^{2n}x^k} left( frac{n(n+1)}{2}x^{n-1}+sum_{k=0}^{n-2}frac{(k+1)(k+2)}{2}left(x^k+x^{2(n-1)-k}right)right)dx$$
An integral to prove that $log(2n+1) ge H_n$
I'm not sure that I would call those expresions 'closed-form'.
– Marc Paul
Jan 25 '16 at 23:48
Yes, they should be regarded as analytic expression because they include infinite summation. Is that what you mean? However, this is the "closest-form" I am aware of for linking $H_n$ and $log(f(n))$
– Jaume Oliver Lafont
Jan 26 '16 at 8:56
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
There is a theory of elementary summation; the phrase generally used is "summation in finite terms." An important reference is Michael Karr, Summation in finite terms, Journal of the Association for Computing Machinery 28 (1981) 305-350, DOI: 10.1145/322248.322255. Quoting,
This paper describes techniques which greatly broaden the scope of what is meant by 'finite terms'...these methods will show that the following sums have no formula as a rational function of $n$:
$$sum_{i=1}^n{1over i},quad sum_{i=1}^n{1over i^2},quad sum_{i=1}^n{2^iover i},quad sum_{i=1}^ni!$$
Undoubtedly the particular problem of $H_n$ goes back well before 1981. The references in Karr's paper may be of some help here.
9
For lazy people like me...
– J. M. is not a mathematician
Jul 20 '11 at 11:20
1
There is also this paper: Michael Karr, "Theory of summation in finite terms", Journal of Symbolic Computation 1 (1985), no. 3, 303–315. MR0849038 (89a:12016)
– lhf
Jul 21 '11 at 22:10
add a comment |
There is a theory of elementary summation; the phrase generally used is "summation in finite terms." An important reference is Michael Karr, Summation in finite terms, Journal of the Association for Computing Machinery 28 (1981) 305-350, DOI: 10.1145/322248.322255. Quoting,
This paper describes techniques which greatly broaden the scope of what is meant by 'finite terms'...these methods will show that the following sums have no formula as a rational function of $n$:
$$sum_{i=1}^n{1over i},quad sum_{i=1}^n{1over i^2},quad sum_{i=1}^n{2^iover i},quad sum_{i=1}^ni!$$
Undoubtedly the particular problem of $H_n$ goes back well before 1981. The references in Karr's paper may be of some help here.
9
For lazy people like me...
– J. M. is not a mathematician
Jul 20 '11 at 11:20
1
There is also this paper: Michael Karr, "Theory of summation in finite terms", Journal of Symbolic Computation 1 (1985), no. 3, 303–315. MR0849038 (89a:12016)
– lhf
Jul 21 '11 at 22:10
add a comment |
There is a theory of elementary summation; the phrase generally used is "summation in finite terms." An important reference is Michael Karr, Summation in finite terms, Journal of the Association for Computing Machinery 28 (1981) 305-350, DOI: 10.1145/322248.322255. Quoting,
This paper describes techniques which greatly broaden the scope of what is meant by 'finite terms'...these methods will show that the following sums have no formula as a rational function of $n$:
$$sum_{i=1}^n{1over i},quad sum_{i=1}^n{1over i^2},quad sum_{i=1}^n{2^iover i},quad sum_{i=1}^ni!$$
Undoubtedly the particular problem of $H_n$ goes back well before 1981. The references in Karr's paper may be of some help here.
There is a theory of elementary summation; the phrase generally used is "summation in finite terms." An important reference is Michael Karr, Summation in finite terms, Journal of the Association for Computing Machinery 28 (1981) 305-350, DOI: 10.1145/322248.322255. Quoting,
This paper describes techniques which greatly broaden the scope of what is meant by 'finite terms'...these methods will show that the following sums have no formula as a rational function of $n$:
$$sum_{i=1}^n{1over i},quad sum_{i=1}^n{1over i^2},quad sum_{i=1}^n{2^iover i},quad sum_{i=1}^ni!$$
Undoubtedly the particular problem of $H_n$ goes back well before 1981. The references in Karr's paper may be of some help here.
edited Apr 21 '16 at 8:15
Martin Sleziak
44.7k8115271
44.7k8115271
answered Jul 20 '11 at 6:11
Gerry MyersonGerry Myerson
146k8147298
146k8147298
9
For lazy people like me...
– J. M. is not a mathematician
Jul 20 '11 at 11:20
1
There is also this paper: Michael Karr, "Theory of summation in finite terms", Journal of Symbolic Computation 1 (1985), no. 3, 303–315. MR0849038 (89a:12016)
– lhf
Jul 21 '11 at 22:10
add a comment |
9
For lazy people like me...
– J. M. is not a mathematician
Jul 20 '11 at 11:20
1
There is also this paper: Michael Karr, "Theory of summation in finite terms", Journal of Symbolic Computation 1 (1985), no. 3, 303–315. MR0849038 (89a:12016)
– lhf
Jul 21 '11 at 22:10
9
9
For lazy people like me...
– J. M. is not a mathematician
Jul 20 '11 at 11:20
For lazy people like me...
– J. M. is not a mathematician
Jul 20 '11 at 11:20
1
1
There is also this paper: Michael Karr, "Theory of summation in finite terms", Journal of Symbolic Computation 1 (1985), no. 3, 303–315. MR0849038 (89a:12016)
– lhf
Jul 21 '11 at 22:10
There is also this paper: Michael Karr, "Theory of summation in finite terms", Journal of Symbolic Computation 1 (1985), no. 3, 303–315. MR0849038 (89a:12016)
– lhf
Jul 21 '11 at 22:10
add a comment |
Harmonic numbers can be represented in terms of integrals of elementary functions:
$$H_n=frac{int_0^{infty} x^n e^{-x} log x ; dx}{int_0^{infty} x^n e^{-x} dx}-int_0^{infty} e^{-x} log x ; dx.$$
This formula could also be used to generalize harmonic numbers to fractional or even complex arguments. These generalized harmonic numbers retain some of their useful properties, for example,
$$H_z=H_{z-1}+frac{1}{z}.$$
9
The Harmonic numbers can also be represented by the integral $H_n = int_0^1 frac{1-x^n}{1-x} dx$
– Rob Bland
Nov 12 '15 at 2:17
add a comment |
Harmonic numbers can be represented in terms of integrals of elementary functions:
$$H_n=frac{int_0^{infty} x^n e^{-x} log x ; dx}{int_0^{infty} x^n e^{-x} dx}-int_0^{infty} e^{-x} log x ; dx.$$
This formula could also be used to generalize harmonic numbers to fractional or even complex arguments. These generalized harmonic numbers retain some of their useful properties, for example,
$$H_z=H_{z-1}+frac{1}{z}.$$
9
The Harmonic numbers can also be represented by the integral $H_n = int_0^1 frac{1-x^n}{1-x} dx$
– Rob Bland
Nov 12 '15 at 2:17
add a comment |
Harmonic numbers can be represented in terms of integrals of elementary functions:
$$H_n=frac{int_0^{infty} x^n e^{-x} log x ; dx}{int_0^{infty} x^n e^{-x} dx}-int_0^{infty} e^{-x} log x ; dx.$$
This formula could also be used to generalize harmonic numbers to fractional or even complex arguments. These generalized harmonic numbers retain some of their useful properties, for example,
$$H_z=H_{z-1}+frac{1}{z}.$$
Harmonic numbers can be represented in terms of integrals of elementary functions:
$$H_n=frac{int_0^{infty} x^n e^{-x} log x ; dx}{int_0^{infty} x^n e^{-x} dx}-int_0^{infty} e^{-x} log x ; dx.$$
This formula could also be used to generalize harmonic numbers to fractional or even complex arguments. These generalized harmonic numbers retain some of their useful properties, for example,
$$H_z=H_{z-1}+frac{1}{z}.$$
edited May 7 '13 at 1:21
answered May 5 '13 at 18:09
Vladimir ReshetnikovVladimir Reshetnikov
24.2k4119231
24.2k4119231
9
The Harmonic numbers can also be represented by the integral $H_n = int_0^1 frac{1-x^n}{1-x} dx$
– Rob Bland
Nov 12 '15 at 2:17
add a comment |
9
The Harmonic numbers can also be represented by the integral $H_n = int_0^1 frac{1-x^n}{1-x} dx$
– Rob Bland
Nov 12 '15 at 2:17
9
9
The Harmonic numbers can also be represented by the integral $H_n = int_0^1 frac{1-x^n}{1-x} dx$
– Rob Bland
Nov 12 '15 at 2:17
The Harmonic numbers can also be represented by the integral $H_n = int_0^1 frac{1-x^n}{1-x} dx$
– Rob Bland
Nov 12 '15 at 2:17
add a comment |
This is probably not what you were looking for (since it isn't in terms of rational or elementary functions), but for the harmonic numbers we have
$$H_n=frac{1}{n!}left[{n+1 atop 2}right]$$
where $left[{n atop k}right]$ are the (unsigned) Stirling numbers of the first kind (page 261 from the book Concrete Mathematics by Graham, Knuth and Patashnik - second edition).
For the generalized harmonic numbers I like this formula - even though it does involve an integral and Riemann zeta...
Maybe you prefer this
add a comment |
This is probably not what you were looking for (since it isn't in terms of rational or elementary functions), but for the harmonic numbers we have
$$H_n=frac{1}{n!}left[{n+1 atop 2}right]$$
where $left[{n atop k}right]$ are the (unsigned) Stirling numbers of the first kind (page 261 from the book Concrete Mathematics by Graham, Knuth and Patashnik - second edition).
For the generalized harmonic numbers I like this formula - even though it does involve an integral and Riemann zeta...
Maybe you prefer this
add a comment |
This is probably not what you were looking for (since it isn't in terms of rational or elementary functions), but for the harmonic numbers we have
$$H_n=frac{1}{n!}left[{n+1 atop 2}right]$$
where $left[{n atop k}right]$ are the (unsigned) Stirling numbers of the first kind (page 261 from the book Concrete Mathematics by Graham, Knuth and Patashnik - second edition).
For the generalized harmonic numbers I like this formula - even though it does involve an integral and Riemann zeta...
Maybe you prefer this
This is probably not what you were looking for (since it isn't in terms of rational or elementary functions), but for the harmonic numbers we have
$$H_n=frac{1}{n!}left[{n+1 atop 2}right]$$
where $left[{n atop k}right]$ are the (unsigned) Stirling numbers of the first kind (page 261 from the book Concrete Mathematics by Graham, Knuth and Patashnik - second edition).
For the generalized harmonic numbers I like this formula - even though it does involve an integral and Riemann zeta...
Maybe you prefer this
edited Dec 25 '11 at 2:42
Srivatsan
20.9k371125
20.9k371125
answered Jul 20 '11 at 9:21
Peter SheldrickPeter Sheldrick
6411339
6411339
add a comment |
add a comment |
$$H_n = frac{binom{(n+1)!+n}{n}-1}{(n+1)!}-(n+1)Biggllfloor frac{binom{(n+1)!+n}{n}-1}{(n+1)(n+1)!}Biggrrfloor $$
That's a remarkable closed form, do you have a source for a proof or more detailed elaboration on how this is derived?
– RocketNuts
May 16 '18 at 23:20
add a comment |
$$H_n = frac{binom{(n+1)!+n}{n}-1}{(n+1)!}-(n+1)Biggllfloor frac{binom{(n+1)!+n}{n}-1}{(n+1)(n+1)!}Biggrrfloor $$
That's a remarkable closed form, do you have a source for a proof or more detailed elaboration on how this is derived?
– RocketNuts
May 16 '18 at 23:20
add a comment |
$$H_n = frac{binom{(n+1)!+n}{n}-1}{(n+1)!}-(n+1)Biggllfloor frac{binom{(n+1)!+n}{n}-1}{(n+1)(n+1)!}Biggrrfloor $$
$$H_n = frac{binom{(n+1)!+n}{n}-1}{(n+1)!}-(n+1)Biggllfloor frac{binom{(n+1)!+n}{n}-1}{(n+1)(n+1)!}Biggrrfloor $$
answered Apr 5 '18 at 10:31
nczksvnczksv
193111
193111
That's a remarkable closed form, do you have a source for a proof or more detailed elaboration on how this is derived?
– RocketNuts
May 16 '18 at 23:20
add a comment |
That's a remarkable closed form, do you have a source for a proof or more detailed elaboration on how this is derived?
– RocketNuts
May 16 '18 at 23:20
That's a remarkable closed form, do you have a source for a proof or more detailed elaboration on how this is derived?
– RocketNuts
May 16 '18 at 23:20
That's a remarkable closed form, do you have a source for a proof or more detailed elaboration on how this is derived?
– RocketNuts
May 16 '18 at 23:20
add a comment |
The following series shows the relationship between the harmonic numbers and the logarithm of odd integers.
$$
log(2n+1)=H_n+sum_{k=1}^{infty}left(sum_{i=-n}^{n}frac{1}{(2n+1)k+i}-frac{1}{k}right)
$$
https://math.stackexchange.com/a/1602945/134791
Equivalently,
$$
H_n=log(2n+1)-sum_{k=1}^{infty}left(sum_{i=-n}^{n}frac{1}{(2n+1)k+i}-frac{1}{k}right)
$$
An integral form is given by
$$H_n=log(2n+1)-int_{0}^{1} frac{x^n(1-x)}{sum_{k=0}^{2n}x^k} left( frac{n(n+1)}{2}x^{n-1}+sum_{k=0}^{n-2}frac{(k+1)(k+2)}{2}left(x^k+x^{2(n-1)-k}right)right)dx$$
An integral to prove that $log(2n+1) ge H_n$
I'm not sure that I would call those expresions 'closed-form'.
– Marc Paul
Jan 25 '16 at 23:48
Yes, they should be regarded as analytic expression because they include infinite summation. Is that what you mean? However, this is the "closest-form" I am aware of for linking $H_n$ and $log(f(n))$
– Jaume Oliver Lafont
Jan 26 '16 at 8:56
add a comment |
The following series shows the relationship between the harmonic numbers and the logarithm of odd integers.
$$
log(2n+1)=H_n+sum_{k=1}^{infty}left(sum_{i=-n}^{n}frac{1}{(2n+1)k+i}-frac{1}{k}right)
$$
https://math.stackexchange.com/a/1602945/134791
Equivalently,
$$
H_n=log(2n+1)-sum_{k=1}^{infty}left(sum_{i=-n}^{n}frac{1}{(2n+1)k+i}-frac{1}{k}right)
$$
An integral form is given by
$$H_n=log(2n+1)-int_{0}^{1} frac{x^n(1-x)}{sum_{k=0}^{2n}x^k} left( frac{n(n+1)}{2}x^{n-1}+sum_{k=0}^{n-2}frac{(k+1)(k+2)}{2}left(x^k+x^{2(n-1)-k}right)right)dx$$
An integral to prove that $log(2n+1) ge H_n$
I'm not sure that I would call those expresions 'closed-form'.
– Marc Paul
Jan 25 '16 at 23:48
Yes, they should be regarded as analytic expression because they include infinite summation. Is that what you mean? However, this is the "closest-form" I am aware of for linking $H_n$ and $log(f(n))$
– Jaume Oliver Lafont
Jan 26 '16 at 8:56
add a comment |
The following series shows the relationship between the harmonic numbers and the logarithm of odd integers.
$$
log(2n+1)=H_n+sum_{k=1}^{infty}left(sum_{i=-n}^{n}frac{1}{(2n+1)k+i}-frac{1}{k}right)
$$
https://math.stackexchange.com/a/1602945/134791
Equivalently,
$$
H_n=log(2n+1)-sum_{k=1}^{infty}left(sum_{i=-n}^{n}frac{1}{(2n+1)k+i}-frac{1}{k}right)
$$
An integral form is given by
$$H_n=log(2n+1)-int_{0}^{1} frac{x^n(1-x)}{sum_{k=0}^{2n}x^k} left( frac{n(n+1)}{2}x^{n-1}+sum_{k=0}^{n-2}frac{(k+1)(k+2)}{2}left(x^k+x^{2(n-1)-k}right)right)dx$$
An integral to prove that $log(2n+1) ge H_n$
The following series shows the relationship between the harmonic numbers and the logarithm of odd integers.
$$
log(2n+1)=H_n+sum_{k=1}^{infty}left(sum_{i=-n}^{n}frac{1}{(2n+1)k+i}-frac{1}{k}right)
$$
https://math.stackexchange.com/a/1602945/134791
Equivalently,
$$
H_n=log(2n+1)-sum_{k=1}^{infty}left(sum_{i=-n}^{n}frac{1}{(2n+1)k+i}-frac{1}{k}right)
$$
An integral form is given by
$$H_n=log(2n+1)-int_{0}^{1} frac{x^n(1-x)}{sum_{k=0}^{2n}x^k} left( frac{n(n+1)}{2}x^{n-1}+sum_{k=0}^{n-2}frac{(k+1)(k+2)}{2}left(x^k+x^{2(n-1)-k}right)right)dx$$
An integral to prove that $log(2n+1) ge H_n$
edited Apr 13 '17 at 12:19
Community♦
1
1
answered Jan 25 '16 at 23:24
Jaume Oliver LafontJaume Oliver Lafont
3,09411033
3,09411033
I'm not sure that I would call those expresions 'closed-form'.
– Marc Paul
Jan 25 '16 at 23:48
Yes, they should be regarded as analytic expression because they include infinite summation. Is that what you mean? However, this is the "closest-form" I am aware of for linking $H_n$ and $log(f(n))$
– Jaume Oliver Lafont
Jan 26 '16 at 8:56
add a comment |
I'm not sure that I would call those expresions 'closed-form'.
– Marc Paul
Jan 25 '16 at 23:48
Yes, they should be regarded as analytic expression because they include infinite summation. Is that what you mean? However, this is the "closest-form" I am aware of for linking $H_n$ and $log(f(n))$
– Jaume Oliver Lafont
Jan 26 '16 at 8:56
I'm not sure that I would call those expresions 'closed-form'.
– Marc Paul
Jan 25 '16 at 23:48
I'm not sure that I would call those expresions 'closed-form'.
– Marc Paul
Jan 25 '16 at 23:48
Yes, they should be regarded as analytic expression because they include infinite summation. Is that what you mean? However, this is the "closest-form" I am aware of for linking $H_n$ and $log(f(n))$
– Jaume Oliver Lafont
Jan 26 '16 at 8:56
Yes, they should be regarded as analytic expression because they include infinite summation. Is that what you mean? However, this is the "closest-form" I am aware of for linking $H_n$ and $log(f(n))$
– Jaume Oliver Lafont
Jan 26 '16 at 8:56
add a comment |
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1
So what is the question?
– Ross Millikan
Jul 20 '11 at 5:07
8
Ross, the second paragraph explicitly and directly states the question. If the first paragraph is too verbose, I can trim it a bit.
– Srivatsan
Jul 20 '11 at 5:25
6
There is an expression which might loosely be called "closed form": $H_n = Psi(n+1) + gamma$, where $Psi$ is the "digamma" function $Psi(x) = frac{d}{dx} ln Gamma(x)$. I don't know how to prove that $Psi$, or $Gamma$ for that matter, is not elementary.
– Robert Israel
Jul 20 '11 at 5:58
In addition to $Psi$, I am not sure if using everyone will agree with using constant $gamma$ which itself doesn't have a nice form. But I am ok with it. (After all, I did allow $e$!)
– Srivatsan
Jul 20 '11 at 6:12
4
FWIW, I consider harmonic numbers as closed forms in themselves, just as I consider $n!$ to be the closed form for $prod_{k=1}^n k$ and $(a)_n$ to be the closed form of $prod_{k=0}^{n-1} (a+k)$...
– J. M. is not a mathematician
Jul 20 '11 at 11:19