Note that arctan x > 0 {\displaystyle \arctan x>0} for x > 0 {\displaystyle x>0} and lim n → ∞ arctan n = π 2 {\displaystyle \lim _{n\to \infty }\arctan n={\frac {\pi }{2}}} .
Applying Limit Comparison Test for a n = arctan n n 2 {\displaystyle a_{n}={\frac {\arctan n}{n^{2}}}} and b n = 1 n 2 {\displaystyle b_{n}={\frac {1}{n^{2}}}} ,
lim n → ∞ arctan n n 2 1 n 2 = lim n → ∞ arctan n = π 2 > 0 {\displaystyle \lim _{n\to \infty }{\frac {\frac {\arctan n}{n^{2}}}{\frac {1}{n^{2}}}}=\lim _{n\to \infty }\arctan n={\frac {\pi }{2}}>0} and the convergence of ∑ n = 1 ∞ 1 n 2 {\displaystyle \sum _{n=1}^{\infty }{\frac {1}{n^{2}}}} implies that ∑ n = 1 ∞ arctan n n 2 < + ∞ . {\displaystyle \sum _{n=1}^{\infty }{\frac {\arctan n}{n^{2}}}<+\infty .}
Therefore the answer is CL.
