Can one put absolute values around the first two function terms in the Euler Maclaurin formula for the Riemann zeta function?

Question

I know by the Euler–Maclaurin formula that this is true for the Riemann zeta function: $$\Re(s)>0:\;\;\;\; \zeta (s)=\lim_{k\to \infty } \, \left|\sum _{n=1}^k \frac{1}{n^s}+\frac{1}{(s-1) k^{s-1}}\right| \;\;\;\;(\ast)$$

My question is if one is allowed to split the absolute value function like this:

$$\Re(s)>0 \;\;\;\; \zeta (s)=\lim_{k\to \infty } \, \left(\left|\sum _{n=1}^k \frac{1}{n^s}\right|-\left|\frac{1}{k^{s-1} (s-1)}\right|\right) \;\;\;\;(\ast \ast)$$

I don't know any mathematical application for such a change other than that $(\ast \ast)$ seems to converge to $0$ slightly faster than the original $(\ast)$ formulation, when $s$ is a Riemann zeta zero.

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(*Mathematica*)
(*start*)
Clear[x, k, j, s];
k = 10000;
s = N[ZetaZero[1], 50];
Abs[Sum[1/n^s, {n, 1, k}] + 1/k^(s - 1)/(s - 1)]
Abs[Sum[1/n^s, {n, 1, k}]] - Abs[1/k^(s - 1)/(s - 1)]
Abs[Sum[1/s^n, {n, 1, k}]] - Abs[1/k^(2 s - 1)/(s - 1)]
(*end*)

Output:

0.004999958472065407135120606606862217064524894
0.000177345880980905486884439449850700223059324