Euler-Maclarurin summation formula and regularization

In this answer, I say

The Euler-Maclaurin Sum formula says that, as a function of $n$, $$ \sum_{k=1}^n\frac{1}{k^z}=\zeta^\ast(z)+\frac{1}{1-z}n^{1-z}+\frac12n^{-z}+O\left(zn^{-1-z}\right)\tag{1} $$ for some $\zeta^\ast(z)$. Note that for $\mathrm{Re}(z)>1$, $\zeta^\ast(z)=\zeta(z)$.

For all $z\in\mathbb{C}\setminus\{1\}$, define $$ \zeta_n(z)=\sum_{k=1}^n\frac{1}{k^z}-\frac{1}{1-z}n^{1-z}\tag{2} $$ Note that each $\zeta_n$ is analytic and equation $(1)$ says that $$ \zeta_n(z)=\zeta^\ast(z)+\frac12n^{-z}+O\left(zn^{-1-z}\right)\tag{3} $$ which says that for $\mathrm{Re}(z)>0$, $$ \lim_{n\to\infty}\zeta_n(z)=\zeta^\ast(z)\tag{4} $$ and the convergence is uniform on compact subsets of $\mathbb{C}\setminus\{1\}$. Thus, the $\zeta^\ast(z)$ defined in $(4)$ is analytic and agrees with $\zeta(z)$ for $\mathrm{Re}(z)>1$. Thus, $\zeta^\ast(z)$ is the analytic continuation of $\zeta(z)$ for $\mathrm{Re}(z)>0$.

In this manner, the Euler-Maclaurin Sum Formula allows us to analytically continue $\zeta(z)$ as far left of $\mathrm{Re}(z)=1$ as we wish.