Answer
Converges
Work Step by Step
Given $$\sum_{n=1}^{\infty}\frac{(\ln n)^{100}}{n^{1.1}} $$ Since for $n\geq1$ \begin{align*} \frac{(\ln n)^{100}}{n^{1.1}}& \leq \frac{\left(n^{0.0001}\right)^{100}}{n^{1.1}}\\ \frac{(\ln n)^{100}}{n^{1.1}} &\leq \frac{n^{0.01}}{n^{1.1}}\\ \frac{(\ln n)^{100}}{n^{1.1}} &\leq \frac{1}{n^{1.09}} \end{align*} Compare with $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^{1.09}}$, a convergent $p- $series $ (p>1)$; then $\displaystyle\sum_{n=1}^{\infty}\frac{(\ln n)^{100}}{n^{1.1}} $ also converges
