This question is related to the following two formulas for $\zeta(s)$.
(1) $\quad\zeta(s)=\frac{1}{1-2^{1-s}}\sum\limits_{n=0}^\infty\frac{1}{2^{n+1}}\sum\limits_{k=0}^n\frac{(-1)^k\binom{n}{k}}{(k+1)^s},\quad s\ne 1\quad\text{(see ref(1) and formula (21) at ref(2))}$
(2) $\quad\zeta(s)=\frac{1}{s-1}\sum\limits_{n=0}^\infty\frac{1}{n+1}\sum\limits_{k=0}^n\frac{(-1)^k\binom{n}{k}}{(k+1)^{s-1}}\qquad\qquad\qquad\text{(see ref(1) and formula (22) at ref(2))}$
Formula (1) above is claimed to converge for $s\ne 1$ at ref(2), but note that $\frac{1}{1-2^{1-s}}$ exhibits a complex infinity at $s=1+i\frac{2\,\pi\,j}{\log(2)}$ where $j\in \mathbb{Z}$ which seems consistent with the convergence claim at ref(1).
Question (1): Is it true that formula (1) converges for $s\ne 1+i\frac{2\,\pi\,j}{\log(2)}$ where $j\in \mathbb{Z}$ versus $s\ne 1$? Or is there an argument about zeros and poles cancelling each other out when formula (1) for $\zeta(s)$ is evaluated at $s=1+i\frac{2\,\pi\,j}{\log(2)}$ where $j\in \mathbb{Z}$ similar to the argument for the convergence of the right side of the functional equation $\zeta(s)=2^s π^{s−1}\sin\left(\frac{π\,s}{2}\right)\,\Gamma(1−s)\,\zeta(1−s)$ at positive integer values of s (e.g. see Using the functional equation of the Zeta function to compute positive integer values)?
Since originally posting question (1) above, I discovered the following Wikipedia article which I believe provides some insight.
Wikipedia Article: Landau's problem with $\zeta(s)=\frac{\eta(s)}{0}$ and solutions
Formula (2) above is claimed to be globally convergent, but seems to exhibit a significant divergence (see Figure (1) below).
Question (2): Is there an error in formula (2), or is there a conditional convergence requirement associated with formula (2) when the outer series is evaluated for a finite number of terms?
ref(1): Wikipedia Article: Riemann zeta function, Representations, Globally convergent series
12/10/2018 Update:
I'm now wondering if formula (2) for $\zeta(s)$ is perhaps only valid for $s\in\mathbb{Z}$.
The following plot illustrates formula (2) for $\zeta(s)$ evaluated for the first $100$ terms.
Figure (1): Illustration of Formula (2) for $\zeta(s)$
The following discrete plot illustrates formula (2) for $\zeta(s)$ minus $\zeta(s)$ where formula (2) is evaluated for the first $100$ terms in blue and the first $1000$ terms in orange.
Figure (2): Discrete Plot of Formula (2) for $\zeta(s)$ minus $\zeta(s)$
Looking at the coefficients of $x_m$ in $$\sum_{k=0}^K 2^{-k-1}\sum_{m=0}^k {k \choose m} x^m = \sum_{k=0} 2^{-k-1}(1+x)^k = \frac{1-2^{-1-K}(1+x)^K}{1-x}$$
as $K \to \infty$ they converge to $1$ boundedly and locally uniformly,
so we find that if $\sum_{n=1}^\infty |a_n| < \infty $ then
$$\sum_{n=1}^\infty a_n = \sum_{k=0}^\infty 2^{-k-1} \sum_{m=0}^k {k \choose m} a_{m+1}$$
With $b_m = (-1)^m a_{m+1}$ then $\sum_{m=0}^k {k \choose m} a_{m+1} = \Delta^k b_m$ is the $k$-th forward difference operator
Summing by parts $l$ times $(1-2^{1-s}) \zeta(s)= \sum_{n=1}^\infty (-1)^{n+1} n^{-s}$, since $\sum_{n=1}^N (-1)^{N+1} = \frac{1+(-1)^{N+1}}{2}$ and $\Delta^k [(-1)^{n+1}n^{-s}] = O(n^{-s-k})$ we obtain that
$$(1-2^{1-s}) \zeta(s) = \sum_{r=0}^{l-1} 2^{-r-1} \sum_{m=0}^r {r \choose m} (-1)^{m} (m+1)^{-s}\\ +2^{-l-1}\sum_{n=1}^\infty (-1)^{n+1}\sum_{m=0}^l {l \choose m} (-1)^{m} (n+m)^{-s}$$
converges absolutely for $\Re(s) > -l+1$.
Letting $a_n = \sum_{m=0}^l {l \choose m} (-1)^{n+m+1} (n+m)^{-s}$ so that $$\sum_{m=0}^k {k \choose m} a_{m+1} = \sum_{m=0}^{l+k} {l+k \choose m} (-1)^{n+m+1} (n+m)^{-s}$$ (forward difference operator $\Delta^{l+k}= \Delta^k \Delta^l$)
we obtain the result
$$(1-2^{1-s}) \zeta(s) = \sum_{r=0}^\infty 2^{-r-1} \sum_{m=0}^r {r \choose m} (-1)^{m} (m+1)^{-s}$$
which is valid for every $s$.
Estimating the rate of convergence isn't obvious, it depends on $Im(s)$.