Does any method of summing divergent series work on the harmonic series?

Question

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?

Answer 1

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.

Answer 2

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

Answer 3

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.

Answer 4

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.

Answer 5

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 $x \nearrow 1$.

Answer 6

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 $\Re(s) = 0$.

Answer 7

Mathematica answers

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

$ \gamma$

See Euler–Mascheroni constant and Borel summation for details.

Answer 8

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]

Answer 9

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!

Answer 10

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.

Answer 11

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

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

Answer 12

Ramanujan summation can be used, and this summation method can be derived by invoking analytic continuation as follows. Consider the partial series $S(N)$ of a divergent series:

$$ S(N) = \sum_{k=1}^N f(k)$$

Let’s split this summation in two parts:

$$ S(N) = \sum_{k=1}^{p-1}f(k) + \sum_{k=p}^N f(k)$$

for some integer $p$. We then we apply the Euler–Maclaurin summation formula to the second summation: $$ S(N) = \sum_{k=1}^{p-1}f(k) + \int_p^{N}f(x) dx + \frac{1}{2}[f(p) + f(N)] + \sum_{r=1}^{\infty} \frac{B_{2r}}{(2r)!}\left[f^{(2r-1)}(N) - f^{(2r-1)}(p)\right]$$

Here the summation over $r$ is usually a divergent asymptotic expansion, it may be written more rigorously as a finite summation plus a remainder term. We can then write $\int_p^N f(x)dx$ in terms of the primitive function as $F(N) - F(p)$. If we also extend the summation over $k$ to $p$ and subtract $f(p)$, we get:

$$ S(N) = \sum_{k=1}^p f(k) + A(N) - A(p)\tag{3}$$

where:

$$ A(u) = F(u) + \frac{f(u)}{2} + \sum_{r=1}^{\infty}\frac{B_{2r}}{(2r)!}f^{(2r-1)}(u)\tag{4}$$

Then we imagine that we could have introduced a parameter in the function $f(x)$, which for some range of the values of that parameter would have made the summation to infinity convergent. Then all the terms that are diverging in the limit of $N$ to infinity in (3) would tend to zero or some constant, and analytically continuing the result back to the value of the parameter that yields the original sum, would have the effect of setting all these diverging terms to zero. We can then do this directly in (3) as follows.

If we denote by $D(u)$ all the terms in $A(u)$ that we're going to set to zero in $D(N)$ and $c$ the constant term in there that we're going to keep, then we can write:

$$\displaystyle A(u) = D(u) + c + \mathcal{o}(1)$$

The constant $c$ can the only come from the primitive function, because after analytic continuation to a domain where the summation is convergent, all the terms in $A(N)$ tend to zero, except possibly $F(N)$. Using that $S(N)$ does not depend on $p$ allows us to take the limit of $p$ to infinity in (3). We can then write:

$$ S(N) = \lim_{p\to\infty}\left[\sum_{k=1}^{p}f(k) - D(p)\right]+ D(N) + \mathcal{o}(1)$$

Deleting the terms in $D(N)$ per the analytic continuation argument and also the terms that tend to zero, gets us to the summation result of:

$$ S = \lim_{p\to\infty}\left[\sum_{k=1}^{p}f(k) - D(p)\right]\tag{5}$$

We see that the constant term $c$ from the primitive function drops out. It's then convenient to define the primitive function such that this constant term is zero. In case of the harmonic series we have $f(k) = \frac{1}{k}$ and $D(p) = \log(p)$, so we have:

$$S = \lim_{p\to\infty}\left[\sum_{k=1}^{p}\frac{1}{k} - \log(p)\right] = \gamma$$

Answer 13

A very general method for evaluating the divergent limit of $f(n)$ arises from considering

$$ f(\lfloor x \rfloor) $$

As a continuous function from $\mathbb{R^+} \rightarrow \mathbb{R} $

If we then form a natural asymptotic expansion as $f(\lfloor x \rfloor) = c_1(x) O_1 (x) + c_2(x) O_2(x) + ... $

Where the $c_i$ are all "eventually bounded" and the $O_i$ are are different asymptotic growth rate. Then you want to compute the average value of the $O(1)$ term to assign your divergent sum. I give some examples below starting from trivial to non trivial:

$$\sum_{n=0}^{\lfloor x \rfloor} \frac{1}{2^n} = 1 + O\left( \frac{1}{x} \right) $$

The $O(1)$ term here is just $1$. So we compute the average value of this as $\frac{1}{1} \int_{0}^{1} 1 dx = 1 $

A little more complex:

$$ \sum_{n=1}^{\lfloor x \rfloor}1 = x -\lbrace x \rbrace $$ And $-\lbrace x \rbrace = O(1)$ so we compute the average value of this as $\int_{0}^{1} -\lbrace x \rbrace dx = -\frac{1}{2}$

The infamous result:

$$ \sum_{n=1}^{\lfloor x \rfloor} n = \frac{1}{2}x^2 + \left(\frac{1}{2} - \lbrace x \rbrace \right)x + \frac{1}{2} \left(\frac{1}{2} - \lbrace x\rbrace \right)^2 - \frac{1}{8} $$

Here the $\frac{1}{2} \left(\frac{1}{2} - \lbrace x\rbrace \right)^2 - \frac{1}{8} $ is the $O(1)$ term and the average value of this is is

$$ \int_{0}^{1} \left(\frac{1}{2} \left(\frac{1}{2} - \lbrace x\rbrace\right)^2 - \frac{1}{8} \right) dx = - \frac{1}{12} $$

By this technique evaluating:

$$ \sum_{n=1}^{\lfloor x \rfloor} \frac{1}{n} = \ln(x) + \gamma + O\left( \frac{1}{x}\right)$$

The standard $\gamma$ result then follows naturally.

Answer 14

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

Answer 15

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.