What would be the value of


Is it $H_{\sqrt{\frac{5}{2}}}$?

Using different definitions of harmonic numbers this question can be computed, but can I use the usual definition of harmonic numbers for any real numbers?

In other words is $$\sum_{k=1}^{n}\frac{1}{k}$$

A useful definition for any real $n$?

When I want to compute the value of $H_{\sqrt{\frac{5}{2}}}$ at Desmos I just define $\sum_{k=1}^{n}\frac{1}{k}$ and I don't put $\sqrt{\frac{5}{2}}$ directly at the upper limit, but I define $n=\sqrt{\frac{5}{2}}$ and the result is exactly what it should be, it seems using a simple substitution will give us the right answer, but generally are we allowed to use the regular definition of harmonic numbers when our $n$ is not necessarily a natural number?

  • $\begingroup$ I would define the $r$th harmonic number for a real number $r$ which is not a negative integer for example by $H_r = \gamma +\psi(r+1)$, where $\gamma$ is the Euler-Mascheroni constant and $\psi$ is the digamma function $\endgroup$ Feb 13, 2020 at 22:29
  • $\begingroup$ Things are even more complicated. Suppose you have a function of $f(n)$ in the domain of real $n$ for which $f(n) = H_n$ for positive integers $n$. Then $g(n) = f(n) + a \sin(\pi n)$ is another extension of $H_n$ to real $n$. Hence you need an additional condition to rule out $a\ne0$. $\endgroup$ Feb 14, 2020 at 7:58

3 Answers 3


Indefinite Sum concept is the answer to your question.

In fact, if a function $f(x)$ is the (forward) difference of a function $F(x)$ $$ f(x) = \Delta _x F(x) = F(x + 1) - F(x) $$ then we say that $F(x)$ is the "antidifference" (or "indefinite sum") of $f(x)$ $$ F(x) = \Delta _x ^{\left( { - 1} \right)} f(x) = \sum\nolimits_x {f(x)} $$

If $ f(x)$ and $F(x)$ are defined over a real, or complex, domain for $x$ then we will have for example $$ \sum\limits_{k = 0}^n {f(x + k)} = \sum\limits_{k = 0}^n {\left( {F(x + 1) - F(x)} \right)} = F(x + n + 1) - F(x) $$

We write the above with a different symbol for the sum as $$ \sum\nolimits_{k = 0}^{\,n} {f(x + k)} = \sum\limits_{k = 0}^{n - 1} {f(x + k)} = F(x + n) - F(x) $$ which can also be written as $$ \sum\nolimits_{k = x}^{\,x + n} {f(k)} = F(x + n) - F(x) $$

The extension to $$ \sum\nolimits_{k = a}^{\,b} {f(k)} = F(b) - F(a) $$ for any real (or complex) $a,b$ inside the domain of definition of $f, F$ is quite natural.

Coming to the harmonic numbers, it is well known that the functional equation of the digamma function is $$ {1 \over x} = \Delta _x \,\psi (x) $$ and it is therefore "natural" to define $$ \bbox[lightyellow] { H_r = \sum\limits_{k = 1}^r {{1 \over k}} = \sum\nolimits_{k = 1}^{\,1 + r} {{1 \over k}} = \psi (1 + r) - \psi (1) = \psi (1 + r) + \gamma } \tag{1}$$ which substantiate M. Janisch's comment

Answering to W. Hintze's comment, please consider how the "indefinite sum" parallels the "indefinite integral - antiderivative" concept.
Similarly to $$ f(x) = {d \over {dx}}F(x)\quad \Leftrightarrow \quad F(x) = \int {f(x)dx} + c $$ we have that $$ f(x) = \Delta \,F(x)\quad \Leftrightarrow \quad F(x) = \sum\nolimits_x {f(x)} + \pi \left( x \right) $$ where now the family of antidifference functions differ by any function $\pi(x)$ , and not by a constant, which is periodic with period (or one of the periods) equal to $1$, as you rightly noticed.

So by "natural extension" I meant to say:
- an extension from integers to reals (and complex) field under the "indefinite sum" concept,
which provides a function $\mathbb C \to \mathbb C$ which fully interpolates $f(n)$;
- in the antidifference family to select the "simpliest / smoothiest" function, same as the Gamma function is selected among the functions satisfying $F(z+1)=zF(z)$ as the only one which is logarithmically convex, or which has the simplest Weierstrass representation, etc.

From the comments I could catch some skepticism as if my answer could just be an "extravagant" personal idea of mine.
That's absolutely not so, the definition in (1) is actually standardly accepted: refer to Wolfram Function Site, and in particular to this section or to the Wikipedia article.
I am just trying to enlight how we can assign a meaning to sums with bounds which are not integral and thus saying that $$ \bbox[lightyellow] { H_{\,1 + \sqrt {5/2} } = \sum\limits_{k = 1}^{\,1 + \sqrt {5/2} } {{1 \over k}} = \sum\nolimits_{k = 1}^{\,\,2 + \sqrt {5/2} } {{1 \over k}} = \psi (\,\,2 + \sqrt {5/2} ) - \psi (1) = 1.7068 \ldots } \tag{2}$$

  • $\begingroup$ I don't believe that the indefinite sum provides an answer to the question. If you believe it does, you should explain that in your answer. Just giving a link is not an adequate response. $\endgroup$
    – Rob Arthan
    Feb 13, 2020 at 21:56
  • $\begingroup$ @RobArthan: your criticism is fully right and accepted: I expanded my answer $\endgroup$
    – G Cab
    Feb 14, 2020 at 0:41
  • $\begingroup$ So finally can I use an irrational index for the lower or upper bound for the sum? $\endgroup$
    – user715522
    Feb 14, 2020 at 6:07
  • $\begingroup$ @ G Cab As to the question of what is a "natural" extension please see my comment to the OP. (and thanks for bringing indefinite sums to my attention). $\endgroup$ Feb 14, 2020 at 8:03
  • $\begingroup$ @Dr.WolfgangHintze: well, the meaning of "natural extension" is .. quite natural, as much as the Gamma function is the natural extension of $(n-1)!$. That the "natural" one does not cover all the possible extensions that's absolutely true: I added some lines to make it clear which is the antidifference family of functions. $\endgroup$
    – G Cab
    Feb 14, 2020 at 17:44

The $n$-th harmonic number $H_n=\sum_{k=1}^n\frac{1}{k}$ admits a nice representation as integral of a finite geometric series which can be generalised in a rather natural way. We have the following representation for $n$ a positive integer: \begin{align*} H_n=\sum_{k=1}^n\frac{1}{k}=\int_0^1\frac{1-t^n}{1-t}dt\tag{1} \end{align*}

The identity (1) is valid since we have \begin{align*} \int_0^1\frac{1-t^n}{1-t}dt&=\int_0^1\left(1+t+\cdots+t^{n-1}\right)\,dt\tag{2}\\ &=\left(t+\frac{1}{2}t^2+\cdots+\frac{1}{n}t^n\right)\bigg|_0^1\tag{3}\\ &=\left(1+\frac{1}{2}+\cdots+\frac{1}{n}\right)-\left(0\right)\tag{4}\\ &=H_n \end{align*}


  • In (2) we apply the finite geometric series formula.

  • In (3) we do the integration.

  • In (4) we evaluate the expression at the upper and lower limit.

The representation (1) indicates a generalisation \begin{align*} H_\color{blue}{r}=\int_0^1\frac{1-t^\color{blue}{r}}{1-t}dt\tag{2} \end{align*} for real values $r$ in fact also for complex values.


  • The formula (2) can be found in the wiki page generalized harmonic numbers.

  • We have an interesting relationship with the Digamma function $\psi(r)$ which has an integral representation \begin{align*} \psi(r+1)=\int_0^1\frac{1-t^r}{1-t}\,dt-\gamma \end{align*} which is strongly related to (2). Here $\gamma$ is the Euler-Mascheroni constant.

  • If we stick at the sigma notation, we often find for real upper limit $r$ the meaning \begin{align*} \sum_{k=1}^r a_k=\sum_{k=1}^{\lfloor r\rfloor}a_k \end{align*} where $\lfloor .\rfloor$ is the floor function.

  • $\begingroup$ You are rigt : in the "traditional" sum notation writing $\sum\limits_{k = r}^s {a_{\,k} } $ actually means $\sum\limits_{k = \,\left\lceil r \right\rceil }^{\left\lfloor s \right\rfloor } {a_{\,k} } $. That's why we need to extend the meaning passing to the antidifference and using a slightly different symbol. $\endgroup$
    – G Cab
    Feb 15, 2020 at 22:05
  • $\begingroup$ @GCab: I think the simpler approach via the integral representation already does the job to generalize $H_n$ appropriately. $\endgroup$ Feb 15, 2020 at 22:11
  • $\begingroup$ Definitely yes, the integral representation does the job very nicely. But I understand the core of the post as being whether and what meaning to assign to a sum with non-integral bounds $\endgroup$
    – G Cab
    Feb 15, 2020 at 22:24
  • $\begingroup$ @GCab: Yes, I know. :-) $\endgroup$ Feb 15, 2020 at 22:25

Your question is "are we allowed to use the regular definition of harmonic numbers when our $n$ is not necessarily a natural number?" The answer is no, since the "regular definition" of a sum is $a_1 + a_2 + ... + a_n$. Here $n$ is the number of summands which can't be a non natural number.

But you can very well give a meaning to the result of a sum. And this result may accept non integers as input.

Take the simpler example of a "sum" $a(\pi) = \text{"}\sum_{k=1}^{\pi} k\text{"}$. This "sum" is not defined, as although the first three summands are 1, 2, and 3, the last 3.14...-th summand is not defined.

Hence the definition starts by calculating the sum up to an integer $n$, and afterwards insert the non integer value for $n$. Here this gives the arithmetic sum $a(n)= \frac{1}{2}n(n+1)$, and hence $a(\pi)= \frac{1}{2}\pi(\pi+1)$.

Now for the harmonic number there is no such simple closed expression for general $n$. But here comes a relation which is valid for natural $n$: $\sum_{k=1}^n \frac{1}{k} = \sum_{k=1}^n \int_0^1 x^{k-1}\,dx=\int_0^1 \sum_{k=1}^n x^{k-1}\,dx=\int_0^1 \frac{1-x^n}{1-x}\,dx$, and now the r.h.s. is valid for also for real $n$.

Your description of the problem gives a good illustration of this procedure. In making the usual definition of the harmonic number $\sum_{k=1}^n \frac{1}{k}$ in your CAS, the system understands that you mean the harmonic number function $H_n$ it has in its "belly", and this is valid for real numbers. If you then insert a value for $n$, be it integer or not, you get the result of this internal function.

  • $\begingroup$ please, let's avoid confusions, do not use the symbol $\sum\nolimits_{k = 1}^{\,\pi } k $ in your argumentation above since it is the symbol adopted (almost standardly) for the antidifference, and $$ \sum\nolimits_{k = 1}^{\,\pi } k = {{\pi \left( {\pi - 1} \right)} \over 2} $$ ! Use instead, $$ \sum\limits_{k = 1}^\pi k = \sum\nolimits_{k = 1}^{\,\pi + 1} k = {{\pi \left( {\pi + 1} \right)} \over 2} $$ And please have a look at the renowned and authoritative "concrete Mathematics" para. 2.6 $\endgroup$
    – G Cab
    Feb 15, 2020 at 21:48
  • $\begingroup$ You might have noticed that I have put the expression in hyphens to avoid confusion. You are free to use any symbol you like, of course, but please avoid forcing others to use your conventions. $\endgroup$ Feb 16, 2020 at 0:06

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