$\sum\limits_{n=1}^{\infty} \frac{\tan^{-1} n}{n}$ diverges.

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?

• I think they are okay. May 27 '18 at 16:45
• Looks good. I fixed some places where you forgot to put a variable in $\tan^{-1}$. May 27 '18 at 16:47
• The exact crossover value in (1) is just $\tan(1)$, no need to plot anything. May 27 '18 at 16:54
• This makes sense; I really should have realized that. Thank you, everyone.
– user465188
May 27 '18 at 16:57
• @Matt.P Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/…
– user
Jun 22 '18 at 20:56

$$\frac{\tan^{-1} n}{n}=\frac1n\left(\frac{\pi}2-\arctan\left(\frac1n\right)\right)\sim \frac{\pi}{2n}+O\left(\frac1{n^2}\right)$$
then the given series diverges by limit comparison test with $\sum \frac1n$.