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”.

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    $\begingroup$ So what is the question? $\endgroup$ – Ross Millikan Jul 20 '11 at 5:07
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    $\begingroup$ Ross, the second paragraph explicitly and directly states the question. If the first paragraph is too verbose, I can trim it a bit. $\endgroup$ – Srivatsan Jul 20 '11 at 5:25
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    $\begingroup$ 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. $\endgroup$ – Robert Israel Jul 20 '11 at 5:58
  • $\begingroup$ 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 $e$!) $\endgroup$ – Srivatsan Jul 20 '11 at 6:12
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    $\begingroup$ 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 $\prod_{k=0}^{n-1} (a+k)$... $\endgroup$ – J. M. isn't a mathematician Jul 20 '11 at 11:19

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 $n$: $$\sum_{i=1}^n{1\over i},\quad \sum_{i=1}^n{1\over i^2},\quad \sum_{i=1}^n{2^i\over i},\quad \sum_{i=1}^ni!$$

Undoubtedly the particular problem of $H_n$ goes back well before 1981. The references in Karr's paper may be of some help here.

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    $\begingroup$ For lazy people like me... $\endgroup$ – J. M. isn't a mathematician Jul 20 '11 at 11:20
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    $\begingroup$ 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) $\endgroup$ – lhf Jul 21 '11 at 22:10

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

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    $\begingroup$ The Harmonic numbers can also be represented by the integral $H_n = \int_0^1 \frac{1-x^n}{1-x} dx$ $\endgroup$ – Rob Bland Nov 12 '15 at 2:17
  • $\begingroup$ @RobBland That does seem to be true when I try it, but I'm having trouble evaluating the integral for arbitrary $n$ and I can't find another source; do you have one? $\endgroup$ – Daniel H Sep 15 '19 at 2:54
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    $\begingroup$ @DanielH For integer $n$ the proof is easy: Use the identity that $\frac{1-x^n}{1-x}$ is equivalent to $1+x+x^2+\cdots+x^{n-1}$ $\endgroup$ – Rob Bland Sep 15 '19 at 21:19
  • $\begingroup$ @RobBland Thanks; I should have been able to figure that out, I use that geometric series thing often enough. $\endgroup$ – Daniel H Sep 16 '19 at 2:33
  • $\begingroup$ @DanielH The integral is by Euler who used the same proof, as shown here. $\endgroup$ – Jam Nov 17 '19 at 20:48

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


$$H_n = \frac{\binom{(n+1)!+n}{n}-1}{(n+1)!}-(n+1)\Biggl\lfloor \frac{\binom{(n+1)!+n}{n}-1}{(n+1)(n+1)!}\Biggr\rfloor $$

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    $\begingroup$ That's a remarkable closed form, do you have a source for a proof or more detailed elaboration on how this is derived? $\endgroup$ – RocketNuts May 16 '18 at 23:20
  • $\begingroup$ @RocketNuts it's probably from the Wikipedia article $\endgroup$ – Geza Kerecsenyi Jan 20 '20 at 20:09
  • $\begingroup$ Hah, what a nice example of a completely useless closed form! $\endgroup$ – darij grinberg Nov 3 '20 at 22:01

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 $\log(2n+1) \ge H_n$

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    $\begingroup$ I'm not sure that I would call those expresions 'closed-form'. $\endgroup$ – Marc Paul Jan 25 '16 at 23:48
  • $\begingroup$ 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 $log(f(n))$ $\endgroup$ – Jaume Oliver Lafont Jan 26 '16 at 8:56

Using this extension of real numbers one can write

$$H_n=\gamma + \operatorname{reg} \ln(\omega_++n)$$

where $\operatorname{reg}$ is taking regular (finite) part of an extended number and $\omega_+$ is a special infinite element $\omega_+=\sum_0^\infty 1=\pi\delta(0)+1/2$.

Moreover, one can write it as

$$H_n=\ln(\omega_++n)+\gamma-\int_1^\infty \frac{dx}x=\ln\left(\frac{\omega_++n}{\omega_+}\right)$$

  • $\begingroup$ Curious why this was downvoted? It doesn't necessarily answer the question but it's incredibly interesting. $\endgroup$ – David Scott Kirby Dec 2 '19 at 19:55
  • $\begingroup$ Dopf! Thank you. $\endgroup$ – David Scott Kirby Dec 2 '19 at 20:43

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