Sorry if this question is slightly out of context.
I've learned that $\zeta(k)=\sum\limits^{\infty}_{z=1} \frac{1}{z^k}$ has meromorphic continuation to $Re(s)>0$.
I know that, from convergence test, $\zeta(s)$ converges when $Re(s)>1$. Also, it is clear that $\zeta(k)$ diverges at $k=0.5$. Meromorphic continuation of $\zeta(s)$ to $Re(s)>0$ means that we would have some meromorphic function $f(s)$ that has the same value with $\zeta(s)$ at $Re(s)>1$ but $f(s)$ can be defined at $Re(s)>0$, with the "compensation" of having some poles. Is this correct?
Yes, but actually no such analytic (or even continuous) continuation exists, since$$\lim_{s\to1^+}|\zeta(s)|=\infty.$$But it turns out that it has an analytic (and even meromorphic) continuation to $\Bbb C\setminus\{1\}$; in particular, it has an analytic continuation to$$\{z\in\Bbb C\mid\operatorname{Re}(z)>0\}\setminus\{1\}.\tag1$$ Actually, it's not hard to prove that such an analytic continuation can be expressed as$$z\mapsto\frac{\sum_{n=1}^\infty(-1)^{n-1}n^{-z}}{1-2^{1-z}}$$for each $z$ in $(1)$ such that $2^{1-z}\ne1$.