Answer
$0$
Work Step by Step
Given: $a_{n}=\frac{(-10)^{n}}{n!}$
A sequence is said to be converged if and only if $\lim\limits_{n \to \infty}a_{n}$ is a finite constant.
Remember that:
$e^{x}=\sum_{n=0}^{\infty}\frac{x^{n}}{n!}$
Therefore,
$\sum_{n=0}^{\infty}\frac{(-10)^{n}}{n!}=e^{-10}$
Apply Divergence Test for Series.
The series $\Sigma a_{n}$ converges if and only if $\lim\limits_{n \to \infty} a_{n}=0$
Since the series $\Sigma a_{n}$ converges where $a_{n}=\frac{(-10)^{n}}{n!}$
We conclude by the divergence test that $\lim\limits_{n \to \infty} a_{n}=\lim\limits_{n \to \infty} \frac{(-10)^{n}}{n!}=0$
Because the limits exists, the sequence converges.
Hence, the given sequence converges to $0$.
