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 $0$.
 A: 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 $0$.
A: You had the right idea to try to use the alternating series test, however you should double check your calculations for the limit of the terms: 
$\dfrac{2^n n!}{5 \cdot 7 \cdots (2n+3)}$ $= \dfrac{3 \cdot 2^nn!}{1 \cdot 3 \cdot 5 \cdots (2n+3)}$ $= \dfrac{3 \cdot 2^nn!}{1 \cdot 3 \cdot 5 \cdots (2n+3)} \cdot \dfrac{2^{n+2}(n+2)!}{2 \cdot 4 \cdots (2n+4)}$ 
$= \dfrac{3 \cdot 2^{2n+2}n!(n+2)!}{(2n+4)!}$ $= \dfrac{3 \cdot 2^{2n+2}(n+2)!^2}{(n+2)(n+1)(2n+4)!}$ $= \dfrac{3 \cdot 2^{2n+2}}{(n+2)(n+1)} \cdot \dbinom{2n+4}{n+2}^{-1}$
$\sim \dfrac{3 \cdot 2^{2n+2}}{(n+2)(n+1)} \cdot \dfrac{\sqrt{\pi(n+2)}}{2^{2n+4}}$ $=  \dfrac{\tfrac{3\sqrt{\pi}}{4}}{(n+1)\sqrt{n+2}} \to 0$ as $n \to \infty$,
Then, by the alternating series test, the series converges. 
In fact, the series converges absolutely, as we have shown that the terms behave like $n^{-3/2}$.
A: For
$\sum \limits _{n=1}^{\infty} (-1)^{n} \dfrac{2^nn!}{5 \cdot 7 \cdot \ldots \cdot (2n+3)}
$,
the ratio of consecutive terms is
$\dfrac{\dfrac{2^{n+1}(n+1)!}{5 \cdot 7 \cdot \ldots \cdot (2n+3)(2n+5)}}{\dfrac{2^nn!}{5 \cdot 7 \cdot \ldots \cdot (2n+3)}}
=\dfrac{2(n+1)}{(2n+5)}
=\dfrac{2n+2}{2n+5}
=1-\dfrac{3}{2n+5}
$
so that the terms decrease
in absolute value
and,
since
$\prod (1-\dfrac{3}{2n+5})
\to 0
$
(because
$\sum \dfrac{3}{2n+5}$
diverges,
the terms go to zero
so the series converges.
