Let $C_n$ be the $n$-th Catalan Number and let $\mathcal{O}_{s,j} = {{2s-j-1}\choose{j}} C_{s-j}^2$. Then we want to consider $\mathcal{E}_s = \sum_{j=0}^{s-1} (-1)^j\mathcal{O}_{s,j}$. We want to show that

$$\frac{\mathcal{E}_s}{C_s^2} \to 0 {\text { as }} s\to \infty.$$

Does anyone have any ideas of how to get this result? We check the first 1000 numerically and I believe it to be true.

For some quick background, $C_s^2$ counts a set of link diagrams and $\mathcal{E}_s$ comes from counting a subset this larger set. So, $\frac{\mathcal{E}_s}{C_s^2}$ is a probability and thus positive for all $s$. I am a topologist, so this kind of question is pretty out of my area and any help would be great.

A few of the things we have tried:

Bounding above by some "simple" functions (like $\frac{2}{s}$), but we can't get a handle on comparing them to $\mathcal{E}_s$.

Bounding above by more complicated function:$$ f(s) = \sum_{j=0}^{s-1} (-1)^j \frac{s^{j}}{j!8^j}$$ It turns out that $f(s) = e^{-s/8} \frac{\Gamma(s+1,-s/8) }{\Gamma(s+1)} $ which is the regualized Gamma function, and we know this goes goes to zero. Again, we can't seem to get a handle on showing $f(s)\geq \frac{\mathcal{E}_s}{C_s^2}$.

- Ratio test for sequences: We tried to show $$\frac{\frac{\mathcal{E}_{s+1}}{C_{s+1}^2} }{\frac{\mathcal{E}_s}{C_s^2} } <1$$ which reduces to the problem $$\frac{\mathcal{E}_{s+1}}{\mathcal{E}_s} <16$$ Here, we thought we could get somewhere but the bounds we need to use to get anywhere end up giving exactly 16, not a value strictly less. Again, for the first 1000, it seems to be no larger than 14, numerically.

So, if anyone happens to know a better way to bound a sum of lots of factorials, I would love to hear about it. Thanks!