# Sum of terms with recurrence relation

I have the following sequence, where $$s$$ is some positive multiple of 4:

$$$$L_n = \begin{cases} \frac{(s-2)!}{2^{s/4-1}(s/2)!(s/4-1)!}, & \text{for n=1} \\ \\ L_{n-1}\cdot\frac{2(s/4-n+1)}{s-2n+1}, & \text{for n=2,3,...\frac{s}{4}} \end{cases}$$$$

I would like to find the value of $$\sum_{k=1}^{s/4}L_k$$.

So far I have tried factoring, but this can't be done repeatedly in any effective way. I'm mainly stumped by how messy these terms are; simplification just doesn't seem possible. I might be willing to settle for upper and lower bounds, provided that they're better than the trivial ones ($$s/4$$ times the smallest and biggest terms). Asymptotic results might be ok as well. Or even just a suggestion of a strategy which might work.

This is not a homework problem, so there's no reason to believe that a solution will be simple, unfortunately.

Edit: I'm currently trying to obtain an upper and lower bound by making estimates on the coefficient on $$L_{n-1}$$. I know that for $$n=2$$ it's very close to $$1/2$$, and then decreases to 0 as $$n$$ gets larger.

• Is $s$ a multiple of $4$? If not, are you summing to $\lfloor{\frac s4}\rfloor$ and how are you defining $(\frac s4)!$? – Rhys Hughes Dec 4 '18 at 19:36
• Sorry, $s$ is a multiple of 4. I'll edit the question to note that. – Alex Dec 4 '18 at 19:37
• @RhysHughes I've also made a small edit of the definition of the terms, since I had mislabeled the index variable initially. – Alex Dec 5 '18 at 18:45

For simplicity put $$r=s/4$$ and $$S(r)=\sum_{k=1}^r L_k$$. For $$n=2\dots, r$$ we have $$L_n<\frac 12L_{n-1}$$, so $$L_n$$ decreases to zero very quickly (and $$L_1). Thus a sum of a few first $$L_k$$ is a good approximation for $$S(r)$$. A computational evidence suggests that
$$S(r)=L_1\left(2-\frac 3{2r}+\frac 3{4r^2}-\frac{T(r)}{r^3}\right),$$
with $$0\le T(r)=O(1)$$.