<s>[INST] 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 $0$.
 [/INST] 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.
</s>