Context:
In this this paper, Juan Arias de Reyna introduces on p12 the following equation for a new power series coefficient:
$$A_m = \lambda_m + \frac{\gamma + \log(4\pi)}{2}-\frac{1}{m}-\frac{1}{m}\sum_{j=2}^{m}(-1)^j{m \choose j} \left(1-\frac{1}{2^j}\right)\zeta(j) \qquad m \in \mathbb{N} \tag{1}$$
Let's multiply (1) by $m$, isolate the last three terms and continue these to $s \in \mathbb{C}$. This gives:
$$f1(s) =s\cdot\frac{\gamma + \log(4\pi)}{2}-1- \sum_{j=2}^{\infty}(-1)^j{s \choose j} \left(1-\frac{1}{2^j}\right)\zeta(j) \qquad s \in \mathbb{C}, \quad \Re(s) > 0 \tag{2}$$
I've found ways to analytically continue both $A_m$ and $\lambda_m$ (= Keiper/Li coefficients) into entire functions and this provided data to assess what the analytic continuation of (2) would look like. After some experiments, I managed to numerically derive the following simple function that appears entire:
$$f2(s) = s\cdot\frac{\gamma + \log(\pi)}{2}-\sum_{j=2}^{\infty}{-s \choose j} \frac{\zeta(j)}{2^j} \qquad s \in \mathbb{C}\tag{3}$$
Question:
Is there a way to prove that the entire function (3) is indeed the full analytic continuation of (2) ?
P.S.
A brief inspection of (3) showed it has at least two real zeros and also a couple of complex zeros, but all with $\Re(s)$ negative.
For $\Re(s)>1$
$$\sum_{j=2}^{\infty}(-1)^j{s \choose j} \left(1-\frac{1}{2^j}\right)\zeta(j)=(1+(-1))^s-1+ s -\sum_{j=2}^{\infty}(-1)^j{s \choose j} \frac{1}{2^j}\zeta(j) $$ $$-1+ s -\sum_{j=2}^{\infty}(-1)^j{s \choose j} \frac{1}{2^j}\zeta(j) $$ The latter is entire as $\frac{1}{2^j}\zeta(j) $ has exponential decay and locally $${s \choose j}= \prod_{m=1}^j (\frac{s+1-m}{m})=O(e^{|s| \sum_{m\le j}1/m})= O(e^{|s|\log j})$$ which means local uniform convergence and analyticity.