# Decaying of a certain ratio of binomial sums

Consider the two sequences $$a(n)=\sum_{k=1}^n\binom{n}k\sum_{j=1}^{k/2}\binom{k}{2j}\frac{(2j)!}{j!}$$ and $$b(n)=\sum_{k=0}^n\binom{n}k^2k!$$

QUESTION. Is this true? $$\frac{a(n)}{b(n)}\longrightarrow 0 \qquad \text{as} \qquad n\rightarrow\infty.$$

It is very much true. We may simplify the first sum by changing the order of summation: $$\sum_k \binom{n}k\binom{k}{2j}=\sum_k\binom{n}{2j}\binom{n-2j}{k-2j}=2^{n-2j}\binom{n}{2j}.$$
Now the summand for $$a(n)$$ is $$2^{n-2j}\binom{n}{2j}\frac{(2j)!}{j!}=2^{n-2j}\frac{n!}{(n-2j)!j!}.$$ The denominator is not less than, say, $$(n/3)!$$, so the fraction is less than $$n!/100^n$$ for large $$n$$, and even if we sum up $$n$$ such fractions we get much less than $$n!.