Convergence of $\sum \limits _{n=1}^{\infty} (-1)^{n} \frac{2^nn!}{5 \cdot 7 \cdot \ldots \cdot (2n+3)}$ $\sum \limits _{n=1}^{\infty} (-1)^{n} \dfrac{2^nn!}{5 \cdot 7 \cdot \ldots \cdot (2n+3)}$
How to check this? I've tried using Leibniz test, it's easy to prove that this one is monotonous, but its limit is rather not $0$.

Rewrite each term into:
$$2\times\frac{4}{5}\times\frac{6}{7}\times\frac{8}{9}\times\frac{10}{11}\times\dots\times\frac{2n}{2n+1}\times\frac{1}{2n+3}<\frac{2}{2n+3}$$
Hence, the limit of the terms go to $0$.
