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It's sort of folklore (as exemplified by this old post at The Everything Seminar) that none of the common techniques for summing divergent series work to give a meaningful value to the harmonic series, and it's also sort of folklore (although I can't remember where I heard this) that the harmonic series is more or less the only important series with this property.

What other methods besides analytic continuation and zeta regularization exist for summing divergent series? Do they work on the harmonic series? And are there other well-known series which also don't have obvious regularizations?

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  • $\begingroup$ Would you include "1+2+4+8+16+...=-1" as a "meaningful" summation? $\endgroup$ – Andrew Critch Oct 29 '09 at 3:52
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    $\begingroup$ Sure: it makes sense both 2-adically and as an example of analytic continuation. $\endgroup$ – Qiaochu Yuan Oct 29 '09 at 4:22
  • $\begingroup$ Did you look at my answer ? I doubt you'll find a simpler one : with a regularization $ e^{-nt}, n^{-t}$ it diverges, with the regularization $n^{-t^3} \cos(t \ln n)$ it converges, why would one be favoured to the other ? $\endgroup$ – reuns Jul 2 '19 at 18:59

13 Answers 13

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One common regularization method that wasn't mentioned in the Everything Seminar post is to take the constant term of a meromorphic continuation. While the Riemann zeta function has a simple pole at 1, the constant term of the Laurent series expansion is the Euler-Mascheroni constant gamma = 0.5772156649...

It is reasonable to claim that most divergent series don't have interesting or natural regularizations, but you could also reasonably claim that most divergent series aren't interesting. Any function with extremely rapid growth (e.g., the Busy Beaver function) is unlikely to have a sum that is regularizable in a natural way.

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    $\begingroup$ When studying the algebraic properties of MZV (eg. in "Algebraic Aspects of Multiple Zeta Values", math/0309425) Michael Hoffman indeed calls treating \zeta(1) as Euler's Gamma a "happy choice" (after Theorem 3.5). $\endgroup$ – Armin Straub Oct 29 '09 at 12:41
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    $\begingroup$ @j0equ1nn The question is essentially asking for a reasonably natural method for obtaining a finite output when the harmonic series is given as input, and taking the constant term of a meromorphic continuation is one way to do so. $\endgroup$ – S. Carnahan Oct 15 '18 at 14:42
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    $\begingroup$ @j0equ1nn The limit of the difference you mention is typically how the Euler-Mascheroni constant $\gamma$ is defined. To prove that the constant term of the continuation of $\zeta$ is equal to $\gamma$ requires a small amount of work. I don't see what is so strange about it - you can think of it as a sub-leading term in an asymptotic expansion. $\endgroup$ – S. Carnahan Oct 17 '18 at 1:03
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    $\begingroup$ @j0equ1nn That’s precisely the “non-rigorous” step, where two limits are interchanged in order to obtain the regularized value. $\endgroup$ – user76284 Jun 27 '19 at 6:31
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    $\begingroup$ @user76284 Okay sure. I think I was kind of tired and cranky when I wrote that. :) I do think you have to be very careful though. Being rigorous is an important part of math. Some of the greats can get away with looking less rigorous in their writing but usually if you dig in, you find they covered all their basis. $\endgroup$ – j0equ1nn Jun 28 '19 at 15:25
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Incidentally, the best text on such questions is Hardy's last book, Divergent Series.

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    $\begingroup$ Is there also an answer to OPs question in this book or do you just comment the question? $\endgroup$ – András Bátkai Apr 15 '18 at 17:51
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    $\begingroup$ @AndrásBátkai It's been most of a decade since we founded mathoverflow --- this question was from its first month of operation, and my answer was perfectly within the style of answers at the time. I agree that, now that the site has evolved and matured, my comment would now be a comment. Hardy's book does provide a number of theorems that should help to answer OP's question, although I don't have the book at hand and so am going from memory. For example, there is no "universal" summation method: summation methods that sum "wild" series necessarily fail to sum "tame" ones. $\endgroup$ – Theo Johnson-Freyd Apr 16 '18 at 0:09
  • $\begingroup$ In particular, I wouldn't be surprised if there were some series that were "too tame" to be summed by any method at all. $\endgroup$ – Theo Johnson-Freyd Apr 16 '18 at 0:10
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    $\begingroup$ Sorry about that. The system gave me this answer to review and I did not look at the date. $\endgroup$ – András Bátkai Apr 16 '18 at 6:51
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    $\begingroup$ No worries. I don't totally understand why "the system" wants very old posts to be reviewed, but your review is correct. $\endgroup$ – Theo Johnson-Freyd Apr 16 '18 at 12:29
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Let $w$ be a state on the quotient C$^*$-algebra $\ell_\infty / c_0$ (bounded sequences quotient out convergent to zero sequences). Then the functional $$ \mathrm{Tr}_w(A) = w ( \{ \frac{1}{\log (1+n)} \sum_{j=1}^n \lambda(n,A) \}_{n=1}^\infty ) $$ is a trace on the ideal of compact operators (on a separable Hilbert space) such that $\mu(n,A) = O(n^{-1})$, $n \geq 1$. Here $\lambda$ denotes the sequence of eigenvalues of the compact operator $A$ ordered so that the sequence of absolute values $| \lambda |$ is a decreasing sequence, and $\mu$ denotes the sequence of singular values (eigenvalues of the absolute value of $A$). If $A_{\mathrm{harmonic}} = \mathrm{diag}(n^{-1})$ (any diagonal operator with the harmonic series as the diagonal) then $\mathrm{Tr}_w(A_{\mathrm{harmonic}})=1$. This is a regularisation of the harmonic series.

Traces on compact operators, thinking of compact operators as noncommutative generalisations of convergent to zero sequences, form summing procedures on these "noncommutative $c_0$ sequences". The trace $\mathrm{Tr}_w$ above is called a Dixmier trace, after the French mathematician Jacques Dixmier who described it in 1968. It has been popularised by Alain Connes in his version of Noncommutative Geometry (Academic Press, 1994). Dixmier traces are not the only traces on the ideal of compact operators such that $\mu(n,A) = O(n^{-1})$, and there exist other traces $\varphi$ such that $\varphi(A_{\mathrm{harmonic}}) = 1$. Dixmier traces generalise the zeta function residue regularisation and the high temperature (or short time) heat kernel regularisation. Thus the zeta function residue regularisation is not the only regularisation possible.

There exist many traces defined on certain ideals besides just the canonical trace on the trace class operators (trace class operators are the noncommutative version of the summable sequences $\ell_1$). Deep results are known about which ideals admit non-trivial traces, which translates as meaning which rates of divergence (of convergent to zero sequences) admit a non-trivial summing procedure. See the book "Singular Traces", De Gruyter 2012 (admission of vested interest: I am one of the authors). The harmonic series fortunately admits a rich non-trivial range of summing procedures. Contrast with $\ell_p$ sequences for $p > 1$ whose associated ideals have no non-trivial traces, and sequences $O(n^{-p})$, $p > 1$, whose associated ideals also have no non-trivial traces.

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As other answers have mentioned, the most "natural" value for the harmonic series seems to be the Euler-Mascheroni constant $\gamma$. The article Euler constant as a renormalized value of Riemann zeta function at its pole by Andrei Vieru says

We believe that Euler constant is not just the "renormalized" value of the Riemann zeta function in 1. In a sense that we shall clarify it is in fact the normal and natural value of zeta of 1. In this paper we first propose a limit definition of a function whose values coincide everywhere with those of the Riemann zeta function, save in 1, where our limit definition yields the Euler constant. Since in the literature one can find more than one way to regularize the value of the zeta function at s=1, we give asymptotic expansions where, by dint of some extended analogies, Euler constant appears to be the true "renormalized" value. As a striking example of such analogies, we propose an expansion of the logarithm function based on Euler constant and on all values of the zeta function at odd positive integers, in which all these presumably irrational numbers are accompanied by Harmonic numbers of corresponding orders. The other aim of this paper is to show how sequences of rationals, often the same, arise in computations related to Dirichlet L-functions. Here, a connection with the Liouville lambda function appears to have been found. Thus we raise the question about the possible usefulness of an extension of the Liouville lambda function to rationals.

One example given in the article (on page 4, equation 11) is

$$\ln \Gamma(x) = -\ln x - \gamma x + \sum_{k \geq 2} \frac{\zeta(k)}{k}(-x)^k$$

where $\Gamma$ is the gamma function and $\zeta$ is the zeta function. If we let $\zeta(1) = \gamma$, this can be simplified to

$$\sum_{k \geq 1} \frac{\zeta(k)}{k} x^k = \ln (-x)!$$

which is, I think, supremely elegant. More generally, we seem to have

$$\sum_{k \geq 1} \zeta(k) k^n x^k = \sum_{k=0}^n \left\{{n+1 \atop k+1}\right\} (-x)^{k+1} \psi^{(k)}(1-x)$$

where $\psi^{(k)}$ is the polygamma function and the brackets indicate Stirling numbers of the second kind.

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Consider the following approach:

\begin{align} \gamma &= \lim_{n \to \infty} \left(\sum_{k=1}^n \frac{1}{k} - \ln n\right) \\ &= \lim_{n \to \infty} \lim_{x \nearrow 1} \left(\sum_{k=1}^n \frac{x^k}{k} - x^n \ln n\right) \\ &= \lim_{n \to \infty} \lim_{x \nearrow 1} \left(\sum_{k=1}^n \frac{x^k}{k} - x^n \ln n\right) + \lim_{x \nearrow 1} 0 \\ &= \lim_{n \to \infty} \lim_{x \nearrow 1} \left(\sum_{k=1}^n \frac{x^k}{k} - x^n \ln n\right) + \lim_{x \nearrow 1} \lim_{n \to \infty} x^n \ln n \\ &\overset{\star}{=} \lim_{x \nearrow 1} \lim_{n \to \infty} \left(\sum_{k=1}^n \frac{x^k}{k} - x^n \ln n\right) + \lim_{x \nearrow 1} \lim_{n \to \infty} x^n \ln n \\ &= \lim_{x \nearrow 1} \left(\lim_{n \to \infty} \left(\sum_{k=1}^n \frac{x^k}{k} - x^n \ln n\right) + \lim_{n \to \infty} x^n \ln n\right) \\ &= \lim_{x \nearrow 1} \lim_{n \to \infty} \left(\sum_{k=1}^n \frac{x^k}{k} - x^n \ln n + x^n \ln n\right) \\ &= \lim_{x \nearrow 1} \lim_{n \to \infty} \sum_{k=1}^n \frac{x^k}{k} \\ &= \lim_{n \to \infty} \lim_{x \nearrow 1} \sum_{k=1}^n \frac{x^k}{k} \\ &= \lim_{n \to \infty} \sum_{k=1}^n \frac{1}{k} \end{align}

where the star indicates the "non-rigorous" step of interchanging limits, which causes the term to diverge rather than converge to $\gamma$, since:

$$\lim_{n \to \infty} \left(\sum_{k=1}^n \frac{x^k}{k} - x^n \ln n\right) = \ln \frac{1}{1-x}$$

for $|x| < 1$, which diverges as $x \nearrow 1$.

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  • $\begingroup$ That's a clever approach to this, and interesting to see the effect of switching the limits. One thing I'm not getting though: doesn't $\lim_{x\nearrow1}(\lim_{n\rightarrow\infty}(x^n\ln n))$ diverge due to $\ln n$ diverging? You have it as though that equals $0$. $\endgroup$ – j0equ1nn Jul 2 '19 at 23:45
  • $\begingroup$ @j0equ1nn $\ln n$ indeed diverges, but $x^n$ goes to 0 faster on $[0, 1)$: desmos.com/calculator/eu9fpspoye $\endgroup$ – user76284 Jul 3 '19 at 0:40
  • $\begingroup$ But isn't $x$ approaching $1$, rather than $0$? $\endgroup$ – j0equ1nn Jul 6 '19 at 3:59
  • $\begingroup$ The limit is approaching 1 from the left, so the actual value at $x=1$ (which does diverge) doesn't matter. What matters are the values "on the way there" or "on that path", i.e. for $x < 1$, all of which go to 0 pointwise. The limit itself, so to speak, doesn't know anything or care about the "actual value" at $x=1$, only what things look like as it approaches $x=1$. Incidentally this is why the order of limits is important. Let me know if this makes sense. $\endgroup$ – user76284 Jul 6 '19 at 4:07
  • $\begingroup$ Okay, I see what you're doing now! Thanks for explaining. I do find this a really clever approach. $\endgroup$ – j0equ1nn Jul 8 '19 at 17:28
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I'm not allowed to post a comment, but in reply to Michael Lugo's post and as a followup to Scott Carnahan, the prime harmonic series can be regularized in analogy with $1 + 1/2 + 1/3 + 1/4 + \ldots$ "$=$" $\gamma$, giving the Mertens constant. See the prime zeta function for more information.

In this case it's not "meromorphic continuation" as the singularity is logarithmic. This leads to the followup question: is there a practical difference, and is there a general theory for the logarithmic (or even more general, e.g. multiply nested logarithmic) case? The prime zeta function has some interesting properties, such as having a natural boundary of analyticity at $\Re(s) = 0$.

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Mathematica answers

Sum[1/n, {n, 1, Infinity}, Regularization -> "Borel"]

$ \gamma$

See Euler–Mascheroni constant and Borel summation for details.

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    $\begingroup$ How did it obtain that Borel sum? $\endgroup$ – user76284 Jun 30 '19 at 4:35
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I have a sneaking suspicion that anything that works on 1/2 + 1/3 + 1/5 + ... will probably also work on the harmonic series, although I certainly don't have any hard reasoning to back this up -- just that it doesn't have nice local properties or nice global properties, much like the harmonic series.

But I sort of hope I'm wrong -- I'd be very interested to see what a regularization for this series looks like!

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There are other sums with no good summation: for example 1+1+1+... Any decent method of summation would yield S=1+S.

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    $\begingroup$ You don't think that 1+1+1+... = -1/2 has some decency? (Of course, that's \zeta(0).) $\endgroup$ – Armin Straub Oct 29 '09 at 12:23
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    $\begingroup$ Hmmm. Perhaps I was too strong there. I tend to assume that a summation method should obey a_1+a_2+...=0+a_1+a_2+..., which zeta regularization does not. But I can't make a strong argument for that assumption. $\endgroup$ – David E Speyer Oct 29 '09 at 12:27
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    $\begingroup$ By the way the proposition that $1+1+1+... = -1/2$ was first made by Euler. $\endgroup$ – Andrew Aug 16 '11 at 20:31
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    $\begingroup$ $\sum_{k=0}^\infty 1 = -1/2$, $\sum_{k=1}^\infty 1 = 1/2$... $\endgroup$ – Anixx Sep 4 '17 at 20:59
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    $\begingroup$ @DavidSpeyer $S = 1+S$ is what you get only if the summation is shift invariant, which is not the case for the zeta summation (that's why it obtains a finite value $-1/2$) $\endgroup$ – reuns Sep 29 '17 at 22:42
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The series 1/2 + 1/3 + 1/5 + ... (the sum of reciprocals of the primes) mentioned by harrison "sums to log log ∞"; more formally,

(1/2 + 1/3 + 1/5 + 1/7 + ... + 1/n) ~ log log n

where ~ has the usual meaning: f(n)~g(n) if lim (n -> infty) f(n)/g(n) = 1.

The nth partial sum of the harmonic series, 1 + 1/2 + 1/3 + ... + 1/n, diverges like log n.

Perhaps sums which diverge "logarithmically fast" are in general problematic, and the harmonic series is just the canonical example of such a series.

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In another question of mine, I found a regulator $f$ such that $f(s,0) = 1$ and

\begin{align} \lim_{\varepsilon \rightarrow 0^+}\lim_{m \rightarrow \infty} \sum_{n=1}^m n^s f(s, n \varepsilon) &= \zeta(-s) \end{align}

for all $s \neq -1$, namely \begin{align} f(s,x) &= \mathrm{e}^{-x}\left(1 - \frac{x}{s+1}\right) \end{align}

Hence we can also regularize the harmonic series as follows: \begin{align} \gamma &= \frac{1}{2} \lim_{\varepsilon \rightarrow 0^+} (\zeta(1+\varepsilon) + \zeta(1-\varepsilon)) \\ &= \frac{1}{2} \lim_{\varepsilon \rightarrow 0^+} \left(\lim_{m \rightarrow \infty} \sum_{n=1}^m n^{-1-\varepsilon} f(-1-\varepsilon, n \varepsilon) + \lim_{m \rightarrow \infty} \sum_{n=1}^m n^{-1+\varepsilon} f(-1+\varepsilon, n \varepsilon) \right) \\ &= \frac{1}{2} \lim_{\varepsilon \rightarrow 0^+} \lim_{m \rightarrow \infty} \left(\sum_{n=1}^m n^{-1-\varepsilon} f(-1-\varepsilon, n \varepsilon) + \sum_{n=1}^m n^{-1+\varepsilon} f(-1+\varepsilon, n \varepsilon) \right) \\ &\overset{\star}{=} \frac{1}{2} \lim_{m \rightarrow \infty} \lim_{\varepsilon \rightarrow 0^+} \left(\sum_{n=1}^m n^{-1-\varepsilon} f(-1-\varepsilon, n \varepsilon) + \sum_{n=1}^m n^{-1+\varepsilon} f(-1+\varepsilon, n \varepsilon) \right) \\ &= \frac{1}{2} \lim_{m \rightarrow \infty} \left(\sum_{n=1}^m n^{-1} f(-1, 0) + \sum_{n=1}^m n^{-1} f(-1, 0) \right) \\ &= \frac{1}{2} \lim_{m \rightarrow \infty} \left(\sum_{n=1}^m n^{-1} + \sum_{n=1}^m n^{-1} \right) \\ &= \lim_{m \rightarrow \infty} \sum_{n=1}^m n^{-1} \\ &= \sum_{n=1}^\infty n^{-1} \\ \end{align}

where the star indicates the non-rigorous step of exchanging limits. Here is the Mathematica code and its plots:

f[s_, x_] := Exp[-x] (1 + a x)
g[s_, t_] := 
 Evaluate@Simplify[
   f[s, t] /. Solve[Integrate[x^s f[s, x], {x, 0, Infinity}] == 0, a],
    Assumptions -> Re[s] > -1]
g[s, t]
Table[{s, 
    Plot[{Zeta[-s], 
      Sum[n^s g[s, n \[Epsilon]], {n, 1, 1000}]}, {\[Epsilon], 0, 1}, 
     Evaluated -> True]}, {s, -4, 4, 1/2}] // TableForm // Quiet
Plot[{EulerGamma, 
  Sum[(n^(-1 + \[Epsilon]) g[-1 + \[Epsilon], n \[Epsilon]] + 
    n^(-1 - \[Epsilon]) g[-1 - \[Epsilon], n \[Epsilon]])/
   2, {n, 1, 1000}]}, {\[Epsilon], 0, 1}, Evaluated -> True]

enter image description here

enter image description here

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  • $\begingroup$ @reuns See the link at the beginning of my answer. $\endgroup$ – user76284 Jul 2 '19 at 0:26
  • $\begingroup$ You are claiming something without proving it then you put too many details. I'm not convinced by your $\lim_{\varepsilon \rightarrow 0^+} \sum_{n=1}^\infty n^s f_s(n \varepsilon) = \zeta(-s)$. Sorry but your linke is a mess. Did you already see any proof of the analytic continuation of $\zeta(s)$ to $\Re(s) > 0$ ? It takes only a few lines $\endgroup$ – reuns Jul 2 '19 at 0:28
  • $\begingroup$ @reuns You need to take the time to read what I linked to. In particular, my derivation is based on Micah’s answer, which follows Terence Tao’s proof. I don’t think it’s appropriate to duplicate it here. $\endgroup$ – user76284 Jul 2 '19 at 0:35
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disclaim: I'm a student major in physics and not in math, with inadequate knowledge of complex analysis, so this answer may have severe mistakes....
It can be understood via Hartogs theorem, at least partially.
Recall that Hartogs extension thm tells us that a complex function with several variables analytic in the connected O\S, where O is open and S is compact, can be extended to S. Then the failure of extension to some subset of S indicates the bad behavior of the whole singularity set.
The main idea is to find some f(z,x) with $f(n,x_0)=H_n$, then try to define the harmonic series as $f(\infty,x_0)$.
Example1: $\Sigma_{n=1}^{\infty} \frac{x^n}{n}$, the function here is some extension of $f(z,x)=\Sigma_{n=1}^{n=z}\frac{x^z}{z}$, which can be special value of Lerch function and yet doesn't matter here, (in the following we may use the $z=\infty$ or $x=\infty$ charts implicitly, so you should apply $z \to 1/z$ and something like that) and O is some neighborhood of $(z=\infty,x=1)$. Yet $f(\infty,x)=-\text{Log}(1-x)$ has a brunch point at $x=1$ and a brunch cut running to $x=\infty$, ie, S is noncompact.
Example2: $\Sigma_{n=1}^{\infty} \frac{n^x}{n}$, this time the value at $z=\infty$ is zeta function with an isolated pole, yet $f(z,1-x)=\zeta(x)-\zeta(z,x)$, the Riemann zeta is analytic for $x\neq1$, and Hurwitz zeta is usually defined for $z>0$ and $x\neq 1$. Roughly the picture is that f is singular at $x=0$, which is removable, and at a family of x-planes located at {z=negative integers} acumulated around z=infty, thus S is noncompact.
Example3: for the $\Sigma_{n=1}^{\infty} \frac{n^x+n^{-x}}{2n}$ regularization appearing in https://math.stackexchange.com/questions/20005/is-it-possible-to-use-regularization-methods-on-the-harmonic-series, the reason is that the singularities are cancelled exactly in pairs and $z=\infty, x=0$ is removable for Hartogs thm.
Posible relation with renormalization: the trick here is to choose proper "conter-terms" cancelling the poles exactly - this is the regularization sheme, just like dimensional regularization, but this leaves constant factors unfixed, then the condition $f(n,x_0)=H_n$ comes to rescue - this is alike the renormalization scheme: we use renormalization conditions to connect the regularized results with true values (experimental values). Yet I think it's differnt from other types of resummation methods since substracting poles will change the value of convergent series.

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$$f_n(t) = \frac{n^{-it-t^3}+n^{it-t^3}}{2}$$

$$\lim_{t \to 0} \sum_{n=1}^\infty f_n(t)\frac{1}{n} = \frac12\lim_{t \to 0} \zeta(1+it+t^3)+\zeta(1-it+t^3)\\ = \frac12\lim_{t \to 0} \frac{1}{it+t^3}+\gamma+O(t)+\frac{1}{-it+t^3}+\gamma+O(t)\\= \frac12\lim_{t \to 0} \frac{1}{it} +O(t)+\frac{1}{-it}+O(t)+2\gamma+O(t)=\gamma$$

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