Why is it impossible to invert the analytic continuation of a Dirichlet series?

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By Mathematica (and the truncated Euler MacLaurin formula) I know that:

$$\zeta(s)=\lim_{k\to \infty } \, \left(\sum _{n=1}^k \frac{1}{n^s}+\frac{1}{(s-1) k^{s-1}}\right) \tag{1}$$ when the real part of $s$ is greater than $0$.

Question:

I would like to understand better why it is impossible to find a Dirichlet series for the reciprocal of the Right Hand Side (RHS) of $(1)$.

In other words I am looking for to invert the reciprocal of the Riemann zeta function:

$$\frac{1}{\zeta(s)}=\frac{1}{\lim_{k\to \infty } \, \left(\sum _{n=1}^k \frac{1}{n^s}+\frac{1}{(s-1) k^{s-1}}\right)} \tag{2}$$ when the real part of $s$ is greater than $0$.

For $\Re(s)>1$ there is the well known relation:

$$\frac{1}{\zeta(s)}=\sum _{n=1}^{\infty} \frac{\mu(n)}{n^s} \tag{3}$$

and the Riemann hypothesis is that $(3)$ converges for $\Re(s)>\frac{1}{2}$.

In a sense I already know the answer to the question. It is because $\frac{1}{(s-1) k^{s-1}}$ is like a constant and $\sum _{n=1}^k \frac{1}{n^s}$ is the Dirichlet series. But can you explain it better to me?