Answer
The series diverges.
Work Step by Step
Use the Ratio Test: $\lim\limits_{n \to \infty}\Big|\frac{a_{n+1}}{a_n}\Big|$. Let $a_n=(-1)^n\frac{(1.1)^n}{n^4}$. Then, $\lim\limits_{n\to \infty}\Bigg|\frac{(-1)^{n+1}(1.1)^{n+1}}{{n+1}^4} \cdot \frac{n^4}{(-1)^n(1.1)^n}\Bigg|$. Because $(1.1)^n$ and $n^4$ are positive for all $n>0$ we can remove the absolute value lines and the alternating terms $(-1)^n$ and $(-1)^{n+1}$. Now, $\lim\limits_{n \to \infty}\Bigg[\frac{(1.1)^{n+1}}{(n+1)^4} \cdot \frac{n^4}{(1.1)^n}\Bigg]= \lim\limits_{n \to \infty}\Bigg[\frac{(1.1)n^4}{(n+1)^4}\Bigg]$ Divide the top and bottom by $n^4$. The limit is then $(1.1) \lim\limits_{n \to \infty}\left[\frac{1}{\frac{(n+1)^4}{n^4}}\right]=(1.1)\lim\limits_{n \to \infty}\left[\frac{1}{(1+\frac{1}{n})^4}\right]$ As $n\to \infty, 1+\frac{1}{n}\to 1$. Therefore, the limit is $(1.1)(1)=1.1>1$, and the series diverges by the Ratio Test.
