Converging/Diverging Series with Factorial I'm looking to determine whether the series converges or diverges. I don't know how to handle this series due to the factorial in the numerator and denominator. Any help is appreciated!!
$$a_n = \frac{ (-1)^{n+1} (n!)^2}{(2n)!}$$
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\begin{align}
&\color{#88f}{\large%
\sum_{n = 1}^{\infty}\pars{-1}^{n + 1}\,{\pars{n}!^{2} \over \pars{2n}!}}
=\sum_{n = 1}^{\infty}\pars{-1}^{n + 1}\,\pars{2n + 1}\,
{\Gamma\pars{n + 1}\Gamma\pars{n + 1} \over \Gamma\pars{2n + 2}}
\\[3mm]&=\sum_{n = 1}^{\infty}\pars{-1}^{n + 1}\,\pars{2n + 1}\,
\int_{0}^{1}t^{n}\pars{1 - t}^{n}\,\dd t
=\int_{0}^{1}\bracks{
\sum_{n = 1}^{\infty}\pars{-1}^{n + 1}\pars{2n + 1}t^{n}\pars{1 - t}^{n}}\,\dd t
\\[3mm]&=\int_{0}^{1}{%
t\pars{t - 1}\pars{t^{2} - t - 3} \over \pars{t^{2} - t - 1}^{2}}\,\dd t
=\int_{-1/2}^{1/2}
{\bracks{\pars{t^{2} - 3/4} + 1}\bracks{\pars{t^{2} - 3/4} - 2}
\over \pars{t^{2} - 3/4}^{2}}\,\dd t
\\[3mm]&=
2\int_{0}^{1/2}\bracks{1 - {1 \over t^{2} - 4} - {2 \over \pars{t^{2} - 4}^{2}}}\,\dd t
\\[3mm]&=\color{#88f}{\large{1 \over 25}\bracks{5 + 4\root{5}{\rm arctanh}\pars{\root{5} \over 5}}} \approx 0.3722
\end{align}
A: Use ratio test. let $a_n = \frac{ (-1)^{n+1} (n!)^2}{(2n)!} $. Then
$$ \left| \frac{ a_{n+1}}{a_n} \right| =  \frac{(n+1)!^2}{(2n+2)! } \cdot \frac{(2n)!}{(n!)^2} = \left( \frac{(n+1) n!}{n!} \right)^2 \cdot \frac{(2n)!}{(2n+2)(2n+1)(2n)!}= \frac{(n+1)^2}{(2n+2)(2n+1)} = \frac{n^2+2n+1}{4n^2+6n+2} \to \frac{1}{4}<1$$
Hence, the series converges
A: The series converges absolutely since: $|\dfrac{a_{n+1}}{a_n}| = \dfrac{(n+1)^2}{(2n+1)(2n+2)} \to \dfrac{1}{4} < 1$ when $n \to \infty$
