What should I use to find the convergence of the series $n\arctan\frac{1}{n^3}$?

Question

I thought of using comparison test and used the other series as $V_n = \frac{1}{n^2}$ . Now using the limit comparsion rule and LH rule to evaluate limit $$\frac{\arctan(\frac{1}{n^3})}{\frac{1}{n^3}}$$ I get the answer as 1 so it should converge as $\frac{1}{n^3}$ converges (p series test) but the book says it is diverging. What am I doing wrong?

Answer 1

By $$\arctan x\leq x$$ for $x>0$, then $$\sum_{n=1}^\infty n\arctan\dfrac{1}{n^3}<\sum_{n=1}^\infty \dfrac{1}{n^2}=\dfrac{\pi^2}{6}$$

Answer 2

Your method is O.K.:

By LH we get $\frac{\arctan x}{x} \to 1$ as $x \to 0$, hence there is $c>0$ such that

$0 < \frac{\arctan x}{x} \le 2$ for $x \in (0,c)$. This gives

$ \arctan(1/n^3) \le 2/n^3$ for $n$ large.