in the sense of infinite series and for an integer 'a' is then correct that
$$ \sum_{n=1}^{\infty}n^{k} = \sum_{n=1}^{a}n^{k}+ \sum_{n=a+1}^{\infty}n^{k} $$
opther that works only when ·$ re(k) > 1 $ ??
i have tried only the case $ a =2 $ and $ k=1$ but does it works for any other values ?
in the zeta regularization spirit
$ \sum_{n=1}^{\infty}n^{k}= \zeta(-k) $ for any value of k excpet k=-1
It generally works. If we define for $a\in \mathbb{N}$ the function
$$\zeta(s,a) := \sum_{n=a+1}^\infty \frac{1}{n^s}$$
for $\operatorname{Re} s > 1$, then $\zeta(\,\cdot\,,a)$ has an analytic continuation to a meromorphic function on the entire plane, with a simple pole at $1$ and no other pole. If we further define the entire function
$$F_a(z) := \sum_{n=1}^a n^{-z},$$
we see that
$$\zeta(s) \equiv F_a(s) + \zeta(s,a)$$
holds for $\operatorname{Re} s > 1$ by the definitions via the series, and the identity theorem then tells us that the identity holds on all of $\mathbb{C}\setminus \{1\}$.