I've come up with what I think are two alternate, valid ways to show that the series $\sum\limits_{n=1}^{\infty} \frac{\tan^{-1}{n}}{n}$ diverges. Hopefully someone can let me know if these hold.
(1) Direct Comparison Test:
For $x$ greater than about $1.557$ or so (an approximation, based on plotting), $\frac{\tan^{-1}{x}}{x} \geq \frac{1}{x}$. So, taking $N = 1$, for $n > N$, we have $\frac{\tan^{-1}{x}}{x} \geq \frac{1}{x} \geq 0$, where the harmonic series diverges. Thus, $\sum\limits_{n=1}^{\infty} \frac{\tan^{-1}{n}}{n}$ also diverges by direct comparison.
(2) Limit-Comparison Test:
Again take our series of comparison to be the harmonic series. We get: \begin{align*} \lim\limits_{n \to \infty} \frac{\frac{\tan^{-1}{n}}{n}}{\frac{1}{n}} & = \lim\limits_{n \to \infty} \tan^{-1} n \\ & = \frac{\pi}{2} \end{align*} Since this ratio is a finite number $\neq 0$, we can conclude that either both series converge or both diverge. Since the harmonic series diverges, $\sum\limits_{n=1}^{\infty} \frac{\tan^{-1} n}{n}$ also diverges.
How do these look?
Thanks in advance.
You derivation are both nice, as an alternative we have that
$$\frac{\tan^{-1} n}{n}=\frac1n\left(\frac{\pi}2-\arctan\left(\frac1n\right)\right)\sim \frac{\pi}{2n}+O\left(\frac1{n^2}\right)$$
then the given series diverges by limit comparison test with $\sum \frac1n$.