Does $$ \sum_{n\ =\ 1}{n!\over \left(\,\sqrt{\,2\,}\, + 1\,\right) \left(\,\sqrt{\,2\,}\, + 2\,\right)\ldots \left(\,\sqrt{\,2\,}\, + n - 1\,\right)\left(\,\sqrt{\,2\,}\, + n\,\right)}\quad $$ converge or diverge ?
I used the D'Alambert criterion, but it gives $D = 1$, and I have no idea what other criterion I could use.
Hint since $$a_{n}=\dfrac{n!}{(\sqrt{2}+1)(\sqrt{2}+2)\cdots(\sqrt{2}+n)}$$ $$\Longrightarrow \dfrac{a_{n}}{a_{n+1}}=\dfrac{n!}{(n+1)!}\cdot\dfrac{(\sqrt{2}+1)(\sqrt{2}+2)\cdots (\sqrt{2}+n+1)}{(\sqrt{2}+1)(\sqrt{2}+2)\cdots(\sqrt{2}+n)}=\dfrac{\sqrt{2}+n+1}{n+1}$$ you can consider Ratio test $$\lim_{n\to\infty}n\left(\dfrac{a_{n}}{a_{n+1}}-1\right)=\lim_{n\to\infty}n\left(\dfrac{\sqrt{2}+n+1}{n+1}-1\right)=\sqrt{2}>1$$ so it is converges