If you read pages 51-55 of the book The Theory Of Functions by Konrad Knopp (Publication date 1947) and you are patient enough to overcome a very bad made digital copy of the book (I do not understand point of making such an archive or was the original book really in such a poor condition? - there are missing parts of mathematical expressions) then the analytic continuation of $\zeta(s)$ can be summarized in two recursive steps:
$$\tag{1}\zeta (s)=\frac{1}{s-1}+\sum _{n=1}^{\infty } n^{-s}-\sum _{n=1}^{\infty } \int _0^1\frac{1}{(n+t)^s}dt, \quad\Re(s)\in (1,\infty)$$ $$\tag{2}\int _0^1\frac{t^q}{(n+t)^{s+q}}dt=\frac{1}{(q+1) (n+1)^{q+s}}+\frac{s+q}{q+1} \int _0^1\frac{t^{q+1}}{(n+t)^{s+q+1}}dt$$
Formula $(2)$ is just an integration by parts.
You can replace integral in $(1)$ with integral in $(2)$ by substituting $q=0$. You get a formula for $\zeta(s)$ valid for $\Re(s)\in (0,\infty)$.
Then substituting $q=1$ in $(2)$ you can replace another integral in $(1)$ and you get formula valid for $\Re(s)\in (-1,\infty)$.
Continuing this recursive process you can get formula for any positive integer $k$ that is valid in the region $\Re(s)\in (-k,\infty)$.
Does the method of Knopp has any connection with the following Mellin transforms?
$$\tag{3}\frac{\zeta (s)}{s}=\int_0^{\infty } \lfloor x\rfloor x^{-s-1} \, dx, \quad\Re(s)\in (1,\infty)$$ $$\tag{4}\frac{\zeta (s)}{s}=\int_0^{\infty } (\lfloor x\rfloor-x) x^{-s-1} \, dx, \quad\Re(s)\in (0,1)$$ $$\tag{5}\frac{\zeta (s)}{s}=\int_0^{\infty } (\lfloor x\rfloor-x+\frac{1}{2}) x^{-s-1} \, dx, \quad\Re(s)\in (-1,0)$$
Both the Knopp method and the Mellin transform increase convergence of formulas by one to the left.
Is there a method that can increase convergence for $\zeta(s)$ formula by $\frac{1}{2}$ to the left?
Particularly I am interested in series for $\zeta(s)$ that would converge in the region $\Re(s)\in (\frac{1}{2},\infty)$ or $\Re(s)\in [\frac{1}{2},\infty)$.
This may not be a complete answer, but hopefully it provides some insight.
I derived the formula
$$\zeta(s)=\underset{K\to\infty}{\text{lim}}\left(\frac{1}{\left(1-2^{1-s}\right) 2^{K+1}}\sum\limits_{n=0}^K \frac{(-1)^n}{(n+1)^s}\sum\limits_{k=0}^{K-n} \binom{K+1}{K-n-k}\right),\quad s\in\mathbb{C}\tag{1}$$
via recursive steps and I believe it is exactly equivalent to the known formula (see Globally convergent series)
$$\quad\zeta(s)=\underset{K\to\infty}{\text{lim}}\left(\frac{1}{1-2^{1-s}}\sum\limits_{n=0}^K\frac{1}{2^{n+1}}\sum\limits_{k=0}^n \binom{n}{k} \frac{(-1)^k}{(k+1)^s}\right)\,,\quad s\in\mathbb{C}\tag{2}$$
for all positive integer values of $K$ and all values of $s$.
The two formulas above have singularities at $s=1+\frac{2\,π\,i\,j}{\log(2)}$ for non-zero integer values of $j$, but I suspect these singularities are removable.
Formulas (1) and (2) above evaluated at a positive integer $K$ converge exactly at all non-positive integer values of $s$ when $|s|\le K$, in other words when $s\in\{0,-1,-2,...,-K\}$, but only converge at other values of $s$ as $K\to\infty$.
Both formulas above are related to formulas for the Dirichlet eta function $\eta(s)=\left(1-2^{1-s}\right) \zeta(s)$ which corresponds to the Dirichlet series
$$\eta(s)=\underset{K\to\infty}{\text{lim}}\left(\sum\limits_{n=1}^K \frac{(-1)^{n-1}}{n^s}\right),\quad\Re(s)>0\tag{3}$$
but the formula
$$\eta(s)=\underset{K\to\infty}{\text{lim}}\left(\sum\limits_{n=0}^K \frac{(-1)^{n}}{(n+1)^s}\right),\quad\Re(s)>0\tag{4}$$ is more closely related to formulas (1) and (2) above.
Assuming $\eta_K(s)$ denotes formula (4) above evaluated using the upper limit $K$, $\eta_K(0)$ evaluates exactly to $\eta(0)-\frac{1}{2}=0$ when $K$ is even and to $\eta(0)+\frac{1}{2}=1$ when $K$ is odd which suggests the analytic continuation
$$\eta(s)=\underset{K\to\infty}{\text{lim}}\left(\frac{\eta_K(s)+\eta_{K+1}(s)}{2}\right),\quad\Re(s)>-1\tag{5}$$
which can be recursively applied $m-1$ times to extend the range of convergence to $\Re(s)>-m$.
Since the recursive process is based on the behavior of the evaluations at non-positive integers, it's highly doubtful to me that the range of convergence of a formula for $\zeta(s)$ can be extended from $\Re(s)>1$ to $\Re(s)>\frac{1}{2}$, but there are formulas related to $\zeta(s)$ which converge for $\Re(s)>\frac{1}{2}$ assuming the Riemann hypothesis such as
$$\quad\frac{\partial\,\log\zeta(s)}{\partial s}=\frac{\zeta'(s)}{\zeta(s)}=-s\left(\frac{1}{s-1}+\int\limits_1^\infty (\psi(x)-x)\, x^{-s-1}\,dx\right),\quad\Re(s)>\frac{1}{2}\tag{6}$$
where
$$\psi(x)=\sum\limits_{n=1}^x \Lambda(n)\tag{7}$$
is the second Chebyshev function and $\Lambda(n)$ is the von Mangoldt functionfunction.