Euler Maclaurin Formula to prove $\zeta(0)=-\frac12$ and $\zeta(-1)=-\frac{1}{12}$ This is a quick question regarding the analytic continuation of the Riemman Zeta function by application of the Euler Maclaurin Formula and the evaluation of $\zeta(0)=-\frac12$ and $\zeta(-1)=-\frac{1}{12}$ .
Choosing $f(x)=\frac{1}{x^s}$ in the second order Euler Maclaurin Formula we obtain
$$\zeta(s)=\frac{1}{s-1}+\frac12+\frac{s}{12}-\frac{s(s+1)}{2}\int_1^\infty\frac{P_2(x)}{x^{s+2}}\,dx \tag{1}$$
Where $P_2(x)$ is the second Periodic Bernoulli polynomial. $(1)$ extends analytically the Riemann zeta function for $-1<\Re(s)$. Obviously, there is a simple pole @ $1$.
If we now set $s \to 0$ in$(1)$ we immediately get that
$$\zeta(0)=-\frac12 \tag{2}$$
Similarly, if we integrate by parts the integral in $(1)$, we get the third order Euler Maclaurin formula that extends  analytically the Riemann zeta function for $-2<\Re(s)$.
$$\zeta(s)=\frac{1}{s-1}+\frac12+\frac{s}{12}+\frac{s(s+1)(s+2)}{2}\int_1^\infty\frac{P_3(x)}{x^{s+3}}\,dx \tag{3}$$
Letting $s \to -1$ in $(3)$  we immediately obtain
$$\zeta(-1)=-\frac{1}{12} \tag{4}$$
Apparently this procedure is all fine because we have extended the domain of the Zeta function to the right of the line $-2$ in the S-complex plane, excluding only the point $s=1$, and now it makes sense to ask what is $\zeta(0)=\, ?$ for instance.
Note also that by letting $s \to 0$ in $(3)$ we again recover that   $\zeta(0)=-\frac12$.
My question is: Is this procedure correct  and valid or there are any caveat that should be taken into account before we perform it, or some additional condition that should be added?

As described in this answer, and computed in this answer, for $\mathrm{Re}(z)\gt-5$, the Euler-Maclaurin Sum Formula says that
$$
\sum_{k=1}^n\frac1{k^z}=\zeta_n(z)+\frac1{1-z}n^{1-z}+\frac12n^{-z}-\frac{z}{12}n^{-1-z}+\frac{z(z+1)(z+2)}{720}n^{-3-z}\tag1
$$
where $\lim\limits_{n\to\infty}\zeta_n(z)=\zeta(z)$.
If we set $z=0$, we get
$$
\sum_{k=1}^n1=\zeta_n(0)+n+\frac12\tag2
$$
If we let $n\to\infty$, we get $\zeta(0)=-\frac12$.
If we set $z=-1$, we get
$$
\sum_{k=1}^nk=\zeta_n(-1)+\frac12n^2+\frac12n+\frac1{12}\tag3
$$
If we let $n\to\infty$, we get $\zeta(-1)=-\frac1{12}$.
If we set $z=-2$, we get
$$
\sum_{k=1}^nk^2=\zeta_n(-2)+\frac13n^3+\frac12n^2+\frac16n\tag4
$$
If we let $n\to\infty$, we get $\zeta(-2)=0$.
If we set $z=-3$, we get
$$
\sum_{k=1}^nk^3=\zeta_n(-3)+\frac14n^4+\frac12n^3+\frac14n^2-\frac1{120}\tag5
$$
If we let $n\to\infty$, we get $\zeta(-3)=\frac1{120}$.
If we set $z=-4$, we get
$$
\sum_{k=1}^nk^4=\zeta_n(-4)+\frac15n^5+\frac12n^4+\frac13n^3-\frac1{30}n\tag6
$$
If we let $n\to\infty$, we get $\zeta(-4)=0$.
