Does Ramanujan summation evaluate the series $\sum \frac{1}{n^s}$ to $\zeta(s)$ or $\zeta(s)-\frac{1}{s-1}$?

Question

On Wikipedia, in the article on Ramanujan summation as well as some related articles, examples of Ramanujan summation of the form $ \sum\frac{1}{n^s}$ are done for various values of $s$ which seem to imply that Ramanujan summation yields $\zeta(s)$.

However other sources such as this longer pedagogical paper on Ramanujan summation, Ramanujan summation of divergent series (PDF) by B Candelpergher, it says for example on page xii in the intro, or equation 1.22 on page 19, and again on page 59, that

$$ \sum^{\mathfrak{R}} \frac{1}{n^{z}}=\zeta(z) - \frac{1}{z-1}. $$

This shorter summary on Ramanujan summation also contains the same formula at the end.

So which is it?

Does

$$ \sum^{\mathfrak{R}} \frac{1}{n^{s}}=\zeta(s) - \frac{1}{s-1}. $$

or is it just

$$ \sum^{\mathfrak{R}} \frac{1}{n^{s}}=\zeta(s) $$

instead?

Are there two different conventions for Ramanujan summation? If so, can someone elucidate their definitions and differences?

Answer 1

this had been a comment, but is now meant as an answer introducing the citation from E. Delabaere, Université d' Angers

I've just skimmed the intro of the Candelspergher-book, and have not much time to go deeper into it. But I see that he says, that the notation $\qquad \displaystyle \sum_{n \ge 0}^\mathcal R \cdots \qquad$ means to have captured the pole of the zeta.
As far as I've understood this, this means that the singularity of the $\zeta(1)$ is removed - and this result is called "Ramanujan sum".

So what he calls the "Ramanujan sum" is actually $\zeta(s)-1/(s-1)$. It seems that it is perhaps a unlucky misnomer. Possibly it were better (like with the "incomplete gamma-function") to write
"The Ramanujan sum of the zeta is the incomplete zeta" or the like,
and thus this should then be called "Ramanujan incomplete sum" to indicate that a completing-term is systematically missing from the sum of the series under discussion. The including of the completion-term would then be called with the common name "Ramanujan-summation"

Then there would be nothing irritating when writing

The "Ramanujan incomplete sum" of the series $1+2+3+4+...$ is $$\sum_{n \ge 1}^{\mathcal R} n = \zeta(-1)-\frac1{-1-1} = -\frac1{12} + \frac12 = \frac5{12}$$ and must be completed by $ - \frac12 $ to arrive at the known value $ - \frac1{12} $ for the zeta-interpretation of this series.

Just my 2 cents...

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update for completeness of my arguments I just include a snippet from E.Delabaeres article on "Ramanujan summation" by the summary of Vincent Puyhaubert, page 86.

• Legend: Here $a(x)$ are the terms of the series, rewritten as when the full series $a_1+a_2+a_3+...$ is expressed in the transformed form $a(1)+a(2)+ a(3)+\cdots $ and the powerseries-representation of $a(x)$ is combined with the Bernoully-numbers (according to the Euler-Maclaurin-formula for this problem)
• The background-colored elements and red ellipses are added by me for pointing to the important terms-of-formula

Answer 2

Yes, it is different conventions and of course I prefer the one which coincides with other regularization methods

$$\sum _{n\ge0}^{\Re} f(n)= -\sum_{n=1}^{\infty} \frac{f^{(n-1)} (0)}{n!} B_n $$

$$\sum _{n\ge1}^{\Re} f(n)= -\sum_{n=1}^{\infty} \frac{f^{(n-1)} (0)}{n!} B_n(1)$$