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...
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
I think that you misread the equation. In the paper, the rhs is $\frac{\log (x)}{m}$ while the expansion at the top of page $85$ is in terms of $\frac m{\log (x)}$.
Anyway, considering
$$-\frac{\zeta '(x)}{\zeta (x)-1}=k$$ around $x=1$, we have
$$\zeta (x)=\frac{1}{x-1}+\gamma -\gamma _1 (x-1)+\frac{1}{2} \gamma _2 (x-1)^2+O\left((x-1)^3\right)$$ So, we know the expansion of $\zeta '(x)$ and lon division gives
$$-\frac{\zeta '(x)}{\zeta (x)-1}=\frac{1}{x-1}+(1-\gamma )+$$ $$\left(2 \gamma _1+1-2 \gamma +\gamma ^2\right)
(x-1)+$$ $$\left(2 \gamma _1+(1-\gamma ) \gamma _1-2 \gamma \gamma _1-\frac{3 \gamma
_2}{2}+1-3 \gamma +3 \gamma ^2-\gamma ^3\right) (x-1)^2+O\left((x-1)^3\right)$$ which is quite good up to $x=5$.
Using series reversion (as @Gary commented)
$$x=1+\frac{1}{k}+\frac{1-\gamma }{k^2}+\frac{2 \left(\gamma _1+1-2 \gamma +\gamma
^2\right)}{k^3}+O\left(\frac{1}{k^4}\right)$$
Edit
If I had to solve the equation for $x$, I would prefer to solve instead
$$-\frac{\zeta (x)-1}{\zeta '(x)}=a \qquad \text{with} \qquad a=\frac 1k$$ which is much better conditioned.
A nonlinear regression gives (the coefficients were made rational), with $R^2 > 0.999999$, as an estimate,
$$x_0=\frac{1-\frac{65 }{92}a-\frac{82 }{273}a^2+\frac{27 }{125}a^3} { 1-\frac{171 }{101}a+\frac{88 }{95}a^2-\frac{16 }{99}a^3}$$
Some results
$$\left(
\begin{array}{ccc}
k & \text{estimate} & \text{solution} \\
1 & 2.91932 & 2.91746 \\
2 & 1.64084 & 1.64232 \\
3 & 1.38780 & 1.38992 \\
4 & 1.27823 & 1.28026 \\
5 & 1.21697 & 1.21881 \\
6 & 1.17784 & 1.17949 \\
7 & 1.15067 & 1.15216 \\
8 & 1.13070 & 1.13205 \\
9 & 1.11541 & 1.11664 \\
10 & 1.10332 & 1.10445
\end{array}
\right)$$
Update
Another possible solution is to build the $[3,3]$ Padé approximant around $x=1$ and, making all coefficients rational have
$$f(x)=-\frac{\zeta (x)-1}{\zeta '(x)}\sim \frac {t+\frac{157 }{1308}t^2+\frac{2 }{309}t^3} {1+\frac{260 }{479}t+\frac{29 }{321}t^2+\frac{1}{224}t^3}=g(x)\qquad \text{where} \qquad t=x-1$$ which is quite good for $1 \leq x \leq 20$. To give an idea
$$\Phi=\int_0^{10} \Big[f(x)-g(x)\Big]^2\,dx=6.35 \times 10^{-6}$$
$$\Phi=\int_0^{20} \Big[f(x)-g(x)\Big]^2\,dx=4.56 \times 10^{-4}$$
Using $g(x)$, we then just need to solve the cubic equation in $t$
$$-1+\left(k-\frac{260}{479}\right) t+\left(\frac{157 k}{1308}-\frac{29}{321}\right)
t^2+\left(\frac{2 k}{309}-\frac{1}{224}\right) t^3=0$$ The discriminant is always negative and using the hyperbolic method, we have the explicit solution for $t(k)$.
Repeating the same calculations as above
$$\left(
\begin{array}{ccc}
k & \text{estimate} & \text{solution} \\
1 & 2.91750 & 2.91746 \\
2 & 1.64232 & 1.64232 \\
3 & 1.38992 & 1.38992 \\
4 & 1.28026 & 1.28026 \\
5 & 1.21881 & 1.21881 \\
6 & 1.17949 & 1.17949 \\
7 & 1.15216 & 1.15216 \\
8 & 1.13205 & 1.13205 \\
9 & 1.11664 & 1.11664 \\
10 & 1.10445 & 1.10445
\end{array}
\right)$$
Best Answer
Ramanujan summation arises out of Euler-Maclaurin summation formula. Ramanujan summation is just (C, 1) summation. (See Cesàro summation)
You can find out easily from Euler-Maclaurin that
$$\sum_{k=1}^{\infty}\frac{1}{k}$$
is not (C, 1) summable.
Follow the method of Ramanujan below (which you can easily follow):
Using Euler-Summation we have
\begin{align*} \zeta(s) & = \frac{1}{s-1}+\frac{1}{2}+\sum_{r=2}^{q}\frac{B_r}{r!}(s)(s+1)\cdots(s+r-2) \\ & \phantom{=} -\frac{(s)(s+1)\cdots(s+q-1)}{q!}\int_{1}^{\infty}B_{q}(x-[x])x^{-s-q} ~dx \end{align*}
$\zeta(s)$ is the Riemann zeta function (Note $s=1$ is pole) . Note that right side has values even for $Re(s)<1$.
For example, putting $s=0$ we get $$\zeta(0)=-\frac{1}{2}.$$
If we put $s=-n$ (n being a positive integer) and $q=n+1$, we see the remainder vanishes and have
$$(n+1)\zeta(-n)=-1+\frac{n+1}{2}+\sum_{r=2}^{n+1}\frac{B_r(-1)^{r-1}}{r!}\binom{n+1}{r}$$
which after
$$\sum_{j=0}^{r}\binom{r+1}{j}B_j=0$$
gives
$$\zeta(-n)=-\frac{B_{n+1}}{n+1}.$$