Let $0 < \alpha < 1$. Show that for $\lambda > 0$ big enough $\sum_{n=1}^\infty \prod_{k=1}^n \frac{1}{\lambda k^{-\alpha} + 1} < \infty$
I think $\lambda = 1$ is enough. You could then use the estimation \begin{align*} \prod_{k=1}^n \frac{1}{ k^{-\alpha} + 1} &\le \prod_{k=1}^n \frac{1}{ n^{-\alpha} + 1} \\ &= \left(\frac{1}{ n^{-\alpha} + 1}\right)^n \\ &= \left(1-\frac{1}{ 1+ n^{\alpha}}\right)^n \\ \end{align*} which looks a bit like the product respresentation of $\exp$.
I've checked it numerically. For small $\alpha$ the expression seems to get quite big, so I'm not 100% sure.