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Does the series $$\sum_{k=0}^{\infty}{\arctan \left(\frac{\sqrt{3}/2}{1/2+k}\right)}$$ diverge ?

This is part of a larger problem, and I am trying to prove the following: we start with the point $\omega = e^{i\frac{\pi}{3}}$ and consider the points $\omega, \omega + 1, \omega + 2,...$ and I want to prove that $Arg(\omega) + Arg(\omega+1) + Arg(\omega+2) + ...$ diverges.

José Carlos Santos
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pangloss
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  • If $f$ is any function such that $f'(0)\ne 0,$ and $a_k$ is a sequence such that $ka_k$ has a finite nonzero limit, then $\sum_{k=1}^{\infty} f(a_k),,\text {diverges}.$ – zhw. Aug 01 '18 at 23:13

2 Answers2

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Note that, by On the arctangent inequality, $\arctan(x)\geq\frac{x}{2}$ for $x\in [0,1]$. Hence, for $k\geq 1$, $${\arctan \left(\frac{\sqrt{3}/2}{1/2+k}\right)}\geq \frac{\sqrt{3}/2}{1+2k}.$$ What may we conclude?

Robert Z
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Note that

$$\arctan \left(\frac{\sqrt{3}/2}{1/2+k}\right)\sim \frac{\sqrt{3}/2}{1/2+k}$$

then refer to limit comparison test with $\sum \frac 1 k$.

user
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