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

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?