The sequence of Stieltjes-constants diverges and thus cannot be summed conventionally. However their signs oscillate (unfortunately non-periodic) and thus I tried Euler- and a version of Noerlund-summation but could not arrive at a convincing result (a first impression is that S is in the near of 0.5, but the partial sums oscillate with any parameter that I can choose).

Q1: what is a meaningful value for the divergent sum $S$ of the Stieltjes-constants?

[update] Meanwhile I worked on the hints from wikipedia concerning the integral in the Borel-summation and could well make my question 2 more precise

I'm trying to make something from the defining exponential generating function for the Stieltjes-constants $$ \zeta(1+z) - {1 \over z} = \sum_{k=0}^\infty (-1)^k { s_k \over k! } z^k \tag1$$ or better with reversed sign of the argument z: $$ \zeta(1-z) + {1 \over z} = \sum_{k=0}^\infty { s_k \over k! } z^k \tag{1.1} $$

because the rhs looks like the inner term in the Borel-summation-method for the Stieltjes-constant, where the integral-term was cancelled. So we might, with wikipedia, say, that $ \zeta(1-z)+{1 \over z}$ is the Borel-transform of my wished sum $ S = \sum_{k=0}^\infty s_k $
So I think, that the correct Borel-sum computation using that $\zeta$-expression is by $$ S = \int_0^\infty \exp(-t) (\zeta(1-tz)+{1\over tz}) dt | _{z=1} \tag2 $$

Feeding this into Pari/GP the integral seems to diverge; we can go to 12 or 13 for the upper bound of the integral to arrive at about S~0.499074922658 but increasing that upper bound further then the numerical integration "begins to diverge" (I got the same value using another method for summation, but again only as partial sum, after which the summation-procedure "begins to diverge") So this would indicate, that the Borel-summation is not sufficient for this, but possibly the value 0.499074... is a legitimate approximation.

It seems, that the alternating sum $A$ can much easier be evaluated to $$ A = \sum_{k=0}^\infty (-1)^k s_k \overset{ \text{ Eulersum}}{=}0.639... \tag3 $$ but this seems not to help yet for the evaluation of the non-alternating series...

Q2: Is formula (2) the correct Borel-summation for the sum S ?
Q2.1: can formula (2) be improved to make the Borel-summation converge?

(I remember that K.Knopp in his monography on series mentioned "iterated Borel-summation" but have no idea how to introduce this here)



(This is more of an extended comment than an answer.)

Starting from the limit definition

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

and using the Abel-Plana summation formula

$$\begin{split}\lim_{n\to\infty}&\left(\sum_{k=m}^n f(k)-\int_m^n f(u)\mathrm du\right)=\\&\frac{f(m)}{2}-\int_{-\infty}^\infty \left(\frac{|t|}{\exp(|2\pi t|)-1}\right)\left(\frac{f(m+it)-f(m-it)}{2it}\right)\mathrm dt\end{split}$$

(see also this and this) gives the following integral representation for the Stieltjes constants:

$$\gamma_k=-2\Im\left(\int_0^\infty \frac{(\log(1+iu))^k}{(1+iu)(\exp(2\pi u)-1)}\mathrm du\right),\qquad k > 0$$

One can then substitute this integral representation into any of the two series considered in the OP. Swapping summation and integration, and then summing the resulting geometric series gives (after separating out $\gamma_0=\gamma$)

$$\begin{align*} S&=\gamma-2\Im\left(\int_0^\infty \frac{\log(1+iu)}{(1+iu)(1-\log(1+iu))(\exp(2\pi u)-1)}\mathrm du\right)\\ A&=\gamma+2\Im\left(\int_0^\infty \frac{\log(1+iu)}{(1+iu)(1+\log(1+iu))(\exp(2\pi u)-1)}\mathrm du\right) \end{align*}$$

I have not been able to find closed forms for these integrals, but I was able to numerically evaluate them in Mathematica. I obtained

$$\begin{align*} S&\approx 0.499074921592010027678427635735828372349293531995548\\ A&\approx 0.639852426254150685730182361093673754144464812998495 \end{align*}$$

  • $\begingroup$ Wow. Thank you very much... This shall take some time to chew on: I was always avoiding the contact with integrals because of lack of elementary introduction in my 70'ies college-math. But perhaps it arrives at the time when I'd invest some effort now - anyway, I'll come back to this only in a couple of days. $\endgroup$ – Gottfried Helms Apr 7 '13 at 20:19
  • $\begingroup$ I just did the other approach (in my other answer) and now tried to get the value for $S$ from the integral using Wolfram alpha - but I do not get a result. Do you have an idea how I could get the formal result or a numerical approximation? (Also Pari/GP cannot integrate this, it seems to converge to a finite value unusable in $S$ if I move the upper bound of integration towards 12,13,20,30 (but cannot go higher much) $\endgroup$ – Gottfried Helms Oct 1 '15 at 13:06
  • $\begingroup$ Have you tried taking the imaginary part before integrating? $\endgroup$ – J. M. is a poor mathematician Dec 12 '15 at 8:45
  • $\begingroup$ of course; but I couldn't make significant progress with this integration so I left it aside a bit later... $\endgroup$ – Gottfried Helms Dec 12 '15 at 12:21
  • $\begingroup$ Then, it is possible that PARI/GP's algorithms are unable to handle this. In Mathematica, I used the double exponential quadrature method of Takahasi and Mori. $\endgroup$ – J. M. is a poor mathematician Dec 12 '15 at 12:27

Hmm, it seems I can answer my own question Q2.1.

First I rewrite in (2.1) the $\zeta$-expression in the integral as the power series with the Stieltjes numbers. Then I append the explicit multiplication with the formal power series for $\exp(-x)$ and finally I integrate termwise. This gives a new power series which seems to be entire, the absolute value of its terms seem to decrease hypergeometrically.

However, to represent the sum of all Stieltjes numbers, it must be evaluated with arguments towards infinity. But it seems, that the function is then not bounded, and is also oscillating in sign - at least using 256 terms of the series and testing for 36 as upper limit of the integral.

So besides that I have now a better tool for the handling of that problem, the question Q1 is still not yet answered...


How about the following, based one of these representations?

\begin{align} \sum_{n=0}^\infty \gamma_n &= \sum_{n=0}^\infty -\frac{\pi}{2(n+1)} \int_{-\infty}^\infty \frac{(\ln(1/2 \pm ix))^{n+1}}{(\cosh\pi x)^2} \,\mathrm{d}x \\ &= -\frac{\pi}{2} \int_{-\infty}^\infty \sum_{n=0}^\infty \frac{(\ln(1/2 \pm ix))^{n+1}}{n+1} \frac{\mathrm{d}x}{(\cosh \pi x)^2} \\ &= -\frac{\pi}{2} \int_{-\infty}^\infty -\ln(1-\ln(1/2 \pm ix)) \frac{\mathrm{d}x}{(\cosh \pi x)^2} \\ &= \frac{\pi}{2} \int_{-\infty}^\infty \frac{\ln(1-\ln(1/2 \pm ix))}{(\cosh \pi x)^2} \,\mathrm{d}x \\ \end{align}

According to WolframAlpha this last integral is about 0.499075, which seems to agree with the value of $S$ in J.M.'s answer.

  • 1
    $\begingroup$ Very nice, ideed. And much concise, thanks! I'll try later to evaluate this using Pari/GP. $\endgroup$ – Gottfried Helms Jul 13 at 23:23
  • $\begingroup$ @GottfriedHelms Thank you. On a tangential note, I think I found a way to compute the regularized sum of all primes using Möbius inversion followed by Abelian means, though I don't have a closed form yet. I'll write it up and post it soon. $\endgroup$ – user76284 Jul 14 at 1:17

A -possibly- even more interesting approach is the following, which uses the concept which I'd called the "Eulerian summation" because it is a summation-procedure for divergent series which uses the (factorially) rescaled matrix of Eulerian numbers (see for that general procedure this article, and also a couple of questions here in MSE and in MO) .
Looking at the $c$'th column ($c$ beginning at $0$) in the involved matrix of coefficients gives the family of power series and their derivatives with the Stieltjesnumbers $s_k$ as coefficients, which all have closed forms involving the zetas at negative arguments and their derivatives (by the standard-definition of the Stieltjes-numbers and the defining formula $(1.1)$ in my original posting/in my question) : $$ \small \begin{array} {r|lllll} c & &\text{sum expression}& &\text{closed form} \\ \hline 0 & \sum_{k=0}^\infty {s_k \over k!} \cdot& 1^k & = \zeta(1-1) +{1\over 1}\\ 1 & \sum_{k=0}^\infty {s_k \over k!} \cdot & \left(2^k - (\binom{k}{1}+1)\cdot 1^k\right) & =\left( \zeta(1-2) +{1\over 2} \right)&- \left((\zeta'(1-1) +{1\over 1}) + (\zeta(1-1) +{1\over 1}) \right)\\ \vdots&& \vdots&\vdots&\vdots \end{array}$$ To make this notation shorter, we'll define the following modification of the $\zeta()$: $$ \tau(s,d) = \left({\zeta^{(d)}(1-s) \over d!} s^d+ \frac 1s \right) \cdot (-1)^d $$ $\qquad \qquad$ and have then the function for the partial sums $$ S_n= \sum_{c=0}^n (-1)^c \tau(n+1-c,c) $$ so that, for instance for $n=2$ we get $$ \begin{array} {} S_2 &=& \tau(3,0) - \tau(2,1) + \tau(1,2) \\ &=& \left( \zeta(1-3)+\frac 13 \right) + \left(\zeta'(1-2)\cdot 2 + \frac 12 \right) + \left(\zeta''(0)/2! + \frac 11 \right)\\ & \approx& 0.499312817978 \end{array}$$ [update] The partial sums arrive the neighbourhood of the final value (as given in the other answer) early, but only to a handful of digits; then they converge furtherly only slowly and then seem to diverge with roughly alternating signs (the sequence of values seems to be Euler-summable, but that's not yet sure).

Anyway, here is a list of convergents for the first few $n$: $$ \small \begin{array}{} n=2&\approx & 0.499312817978 \\ n=3&\approx & 0.499127252795 \\ n=5&\approx & 0.499105696659 \\ n=6&\approx & 0.499104688195 \\ n=8&\approx & 0.499079388211 \\ \cdots\\ \to \infty & & 0.499074921592010 \ldots \\ \cdots\\ n=7&\approx & 0.499026624450 \\ n=4&\approx & 0.498990483194 \\ \end{array}$$


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