Answer
$\displaystyle\sum_{n=1}^{\infty}(\arctan n)^n$ is divergent.
Work Step by Step
Let\[\sum_{n=1}^{\infty}a_n=\sum_{n=1}^{\infty}(\arctan n)^n\]
\[\Rightarrow a_n=(\arctan n)^n \]
\[\Rightarrow a_n^{\frac{1}{n}}=[(\arctan n)^n]^{\frac{1}{n}}=\arctan n \]
\[\lim_{n\rightarrow\infty}a_n^{\frac{1}{n}}=\lim_{n\rightarrow\infty}\arctan n\]
\[\lim_{n\rightarrow\infty}a_n^{\frac{1}{n}}=\frac{\pi}{2}>1\]
By root test,
$\displaystyle\sum_{n=1}^{\infty}(\arctan n)^n$ is divergent.
