# Prove the absolute convergence of the series?

To prove that sum_{n=1}^{infty}((-100)^{n})/(n!) converges absolutely, we must show that sum_{n=1}^{infty}|((-100)^{n})/(n!)|=sum_{n=1}^{infty}(100^{n})/(n!) converges.
This is best done by the Ratio Test. Let a_{n}=100^{n}/(n!). Then a_{n+1}=(100*100^{n})/((n+1)*n!) and ${a}_{n + 1} / {a}_{n} = \frac{100}{n + 1} \to 0 = L$ as $n \to \infty$.
Since $L < 1$, it follows that sum_{n=1}^{infty}a_{n}=sum_{n=1}^{infty}(100^{n})/(n!) converges. Hence, sum_{n=1}^{infty}((-100)^{n})/(n!) converges absolutely.