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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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Would you include "1+2+4+8+16+...=-1" as a "meaningful" summation? – Andrew Critch Oct 29 '09 at 3:52
Sure: it makes sense both 2-adically and as an example of analytic continuation. – Qiaochu Yuan Oct 29 '09 at 4:22
up vote 12 down vote accepted

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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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). – Armin Straub Oct 29 '09 at 12:41

Incidentally, the best text on such questions is Hardy's last book, Divergent Series.

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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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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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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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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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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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You don't think that 1+1+1+... = -1/2 has some decency? (Of course, that's \zeta(0).) – Armin Straub Oct 29 '09 at 12:23
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. – David Speyer Oct 29 '09 at 12:27
By the way the proposition that $1+1+1+... = -1/2$ was first made by Euler. – Andrew Aug 16 '11 at 20:31

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, 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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