Answer
converges to $0$.
Work Step by Step
As we know that a sequence converges when $\lim\limits_{n \to \infty}a_n$ exists.
Consider $a_n= \Sigma_{n=1}^{\infty}\dfrac{(-4)^n}{n!}$
$\Sigma_{n=1}^{\infty}\dfrac{x^n}{n!}=e^x$
Thus,
$a_n= \Sigma_{n=1}^{\infty}\dfrac{(-4)^n}{n!}=e^{-4}$
Apply limits to both sides.
$\lim\limits_{n \to \infty}a_n=\lim\limits_{n \to \infty} e^{-4}$
Since, the term does not consist any n-term so the sequence must converge to $0$.
