Convergence of the series $\sum n!/(n^2+3)$

Question

How can we test if this series diverges/converges?

$$\sum_{n=1}^\infty\frac{n!}{n^2+3}$$

I tried D'Alembert's principle and tried to do $\frac{a_{n+1}}{a_n}$ but I'm stuck. Any help?

Answer 1

get the lower bound on the summand and show it doesn't tend to $0$, like $a_n >(n-3)! \to \infty$.

Answer 2

For $n\geqslant 2$, $n!\geqslant n(n-1)(n-2)\geqslant (n-2)^3$, hence for $n\geqslant 4$, using the inequalities $n-2\geqslant n/2$ and $n^2\geqslant 1$, $$\frac{n!}{n^2+3}\geqslant\frac{(n-2)^3}{n^2+1}\geqslant \frac{n^3}{2(n^2+1)}\geqslant \frac{n^3}{2\cdot 2n^2}.$$

Answer 3

BIG HINT:

$$\lim\limits_{n\to\infty}\frac{n!}{n^2+3}=\infty$$