Answer
Absolutely convergent
Work Step by Step
Let $a_n=\frac{n^{10}}{(-10)^{n+1}}$. Then $\lim\limits_{n \to \infty}\left|\frac{a_{n+1}}{a_n}\right|= \lim\limits_{n \to \infty}\left|\frac{\frac{(n+1)^{10}}{(-10)^{(n+1)+1}}}{\frac{n^{10}}{(-10)^{n+1}}}\right|=\lim\limits_{n \to \infty}\left|\frac{(n+1)^{10}}{(-10)^{n+2}}\cdot\frac{(-10)^{n+1}}{n^{10}}\right|\lim\limits_{n \to \infty}\left|\frac{(n+1)^{10}}{n^{10}}\cdot\frac{(-10)^{n+1}}{(-10)^{n+2}}\right|=\lim\limits_{n \to \infty}\left|\frac{(n+1)^{10}}{n^{10}}\cdot\frac{1}{-10}\right| =$
$$\frac{1}{10}\cdot\lim\limits_{n \to \infty}\left(\frac{n+1}{n}\right)^{10}=$$
$$\frac{1}{10}\cdot\lim\limits_{n \to \infty}\left(1 +\frac{1}{n}\right)^{10}= \frac{1}{10} \cdot1 =\frac{1}{10}.$$
Since $\lim\limits_{n \to \infty}\left|\frac{a_{n+1}}{a_n}\right|=\frac{1}{10} < 1$, the series $\displaystyle{\sum_{n=1}^{\infty}\frac{n^{10}}{(-10)^n}}$ is absolutely convergent by the Ratio Test.