Showing divergence of the series with terms $\frac{1}{2}-\frac{1}{\pi}\tan^{-1}\left(n\right)$

Question

I'm trying to show that $\sum_{n=1}^{\infty}\left(\frac{1}{2}-\frac{1}{\pi}\tan^{-1}\left(nc\right)\right)$ diverges. Wolframalpha claims it divergence by the comparison test but I have no idea what is it comparing to. The ratio/root tests are both inconclusive and the integral test gets really messy. I would appreciate it if anyone could give me a pointer on how exactly one tries to prove divergence in a case like this. I obviously tried comparing to the harmonic series but its terms are unfortunately always above the terms of my series.

Edit:I forgot there is also the limit-comparison test and that one works with the harmonic series. On a more general note though, are there any rules of thumb one should employ when trying to analyze divergence/convergence of a series?

Answer 1

Put $$L = \lim_{n\to\infty} n(1/2 - (1/\pi)\tan^{-1}(nc)) = \lim_{t\downarrow 0} {1/2 - (1/\pi)\tan^{-1}(c/t)\over t}$$ This is a $0/0$ indeterminate form, so apply L'Hospital to get $$L = {1\over \pi}\lim_{t\downarrow 0} {c/t^2\over 1 + (c/t)^2} = = {1\over \pi} \lim_{t\downarrow 0}{c\over t^2 + c^2} {1\over \pi}$$

We have just shown $$1/2 - (1/\pi)\tan^{-1}(cn)\sim {1\over cn\pi}. $$ By the limit comparison test, your series diverges. This, provided $c\not= 0$.