Do harmonic numbers have a “closed-form” expression?












42














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”.










share|cite|improve this question




















  • 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
















42














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”.










share|cite|improve this question




















  • 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














42












42








42


21





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”.










share|cite|improve this question















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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








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














  • 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










5 Answers
5






active

oldest

votes


















32





+50









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.






share|cite|improve this answer



















  • 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





















18














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,
$$H_z=H_{z-1}+frac{1}{z}.$$






share|cite|improve this answer



















  • 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





















15














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






share|cite|improve this answer































    2














    $$H_n = frac{binom{(n+1)!+n}{n}-1}{(n+1)!}-(n+1)Biggllfloor frac{binom{(n+1)!+n}{n}-1}{(n+1)(n+1)!}Biggrrfloor $$






    share|cite|improve this answer





















    • 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





















    1














    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 $log(2n+1) ge H_n$






    share|cite|improve this answer























    • 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













    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f52572%2fdo-harmonic-numbers-have-a-closed-form-expression%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    5 Answers
    5






    active

    oldest

    votes








    5 Answers
    5






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    32





    +50









    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.






    share|cite|improve this answer



















    • 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


















    32





    +50









    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.






    share|cite|improve this answer



















    • 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
















    32





    +50







    32





    +50



    32




    +50




    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.






    share|cite|improve this answer














    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.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    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
















    • 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













    18














    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,
    $$H_z=H_{z-1}+frac{1}{z}.$$






    share|cite|improve this answer



















    • 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


















    18














    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,
    $$H_z=H_{z-1}+frac{1}{z}.$$






    share|cite|improve this answer



















    • 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
















    18












    18








    18






    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,
    $$H_z=H_{z-1}+frac{1}{z}.$$






    share|cite|improve this answer














    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,
    $$H_z=H_{z-1}+frac{1}{z}.$$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    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
















    • 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













    15














    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






    share|cite|improve this answer




























      15














      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






      share|cite|improve this answer


























        15












        15








        15






        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






        share|cite|improve this answer














        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







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 25 '11 at 2:42









        Srivatsan

        20.9k371125




        20.9k371125










        answered Jul 20 '11 at 9:21









        Peter SheldrickPeter Sheldrick

        6411339




        6411339























            2














            $$H_n = frac{binom{(n+1)!+n}{n}-1}{(n+1)!}-(n+1)Biggllfloor frac{binom{(n+1)!+n}{n}-1}{(n+1)(n+1)!}Biggrrfloor $$






            share|cite|improve this answer





















            • 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


















            2














            $$H_n = frac{binom{(n+1)!+n}{n}-1}{(n+1)!}-(n+1)Biggllfloor frac{binom{(n+1)!+n}{n}-1}{(n+1)(n+1)!}Biggrrfloor $$






            share|cite|improve this answer





















            • 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
















            2












            2








            2






            $$H_n = frac{binom{(n+1)!+n}{n}-1}{(n+1)!}-(n+1)Biggllfloor frac{binom{(n+1)!+n}{n}-1}{(n+1)(n+1)!}Biggrrfloor $$






            share|cite|improve this answer












            $$H_n = frac{binom{(n+1)!+n}{n}-1}{(n+1)!}-(n+1)Biggllfloor frac{binom{(n+1)!+n}{n}-1}{(n+1)(n+1)!}Biggrrfloor $$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            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




















            • 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













            1














            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 $log(2n+1) ge H_n$






            share|cite|improve this answer























            • 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


















            1














            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 $log(2n+1) ge H_n$






            share|cite|improve this answer























            • 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
















            1












            1








            1






            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 $log(2n+1) ge H_n$






            share|cite|improve this answer














            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 $log(2n+1) ge H_n$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            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




















            • 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




















            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.





            Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


            Please pay close attention to the following guidance:


            • Please be sure to answer the question. Provide details and share your research!

            But avoid



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f52572%2fdo-harmonic-numbers-have-a-closed-form-expression%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            Popular posts from this blog

            1300-talet

            1300-talet

            Has there ever been an instance of an active nuclear power plant within or near a war zone?