Determine $\sum _{n=1}^{\infty }\dfrac{1}{(n!)!}$ conv or div?
I tried applying the root test but I'm not sure how to evaluate this double factorial limit? Apparently it converges. I also don't know what other test I could apply?
$$\lim_{n\to\infty} |\dfrac{((n+1)!)!}{(n!)!}|$$
$$\dfrac{1}{(n!)!}<\dfrac{1}{n!}\qquad\forall\,n\ge2$$ (because $(n!)!=(n!)(n!-1)(n!-2)\cdots 2\cdot1>n!$ when $n\ge 2$) where $\sum\dfrac{1}{n!}$ converges by the ratio test: $$\lim_{n\to\infty}\left\vert\dfrac{a_{n+1}}{a_{n}}\right\vert=\lim_{n\to\infty}\dfrac{n!}{(n+1)!}=\lim_{n\to\infty}\dfrac{1}{n+1}=0<1$$