Context:
While working on Keiper's $\tau_s$-function, I found that it could be split into a smooth first term and an oscillating second term:
$$\tau_s=\sum_{k=1}^{\infty} \left( {s+k-1\choose k-1}\cdot \frac{\zeta(k+1)}{(-2)^{k+1}}- {-s+k-1\choose k-1} \sigma_{k+1}^*\right)$$
with $\sigma_k^*$ defined here. I found that numerically the first term appears asymptotic to $\dfrac{1}{2s}$ for $\Re(s) \gt 0$.
Question:
With $s \in \mathbb{C}, \Re(s) > 0$, is the following asymptotic true ?
$$\sum_{k=1}^{\infty} {s+k-1\choose k-1}\cdot \frac{\zeta(k+1)}{(-2)^{k+1}}\asymp \frac{1}{2s} \qquad \text{for} \quad \Re(s) \rightarrow \infty$$
Use $${n+\alpha\choose n} = (-1)^n {-\alpha-1\choose n}$$ Then $$\sum_{k=0}^{\infty} {s+k\choose k}\frac{\zeta(k+2)}{(-2)^{k+2}} = \sum_{n\ge 1} \sum_{k=0}^{\infty} {-s-1\choose k} (2n)^{-k-2} = \sum_{n\ge 1}(2n)^{-2}(1+\frac1{2n})^{-s-1}$$ $$\sim \frac12 \int_1^\infty x^{-2} (1+\frac1{x})^{-s-1}dx=\frac12 \int_0^1 (1+y)^{-s-1}dy=\frac{1-2^{-s}}{2s}$$