Sum of angles in this spiral is diverging.

Question

Suppose the angles are $\theta_n=\arctan(n^{-1/2}).$ I want to show that $$\sum_{n=1}^{\infty}\theta_n\rightarrow \infty.$$

Here is what I did: I want to find a series $s_n$ that I know is diverging and I want to show that the value of its elements is decreasing faster than the original series. Then my original series must diverge.

Let $n\in \mathbb{R^+}$. Then $$\frac{d}{dn}\theta_n=-\frac{1}{2}\frac{1}{n^{-1/2}}\frac{1}{n+1}.$$

After some large enough $N$ it is true that $$-\frac{1}{2}\frac{1}{N^{-1/2}}\frac{1}{N+1}>-\frac{1}{N^2}.$$

Now $$s_n=\frac{1}{n} \Rightarrow\frac{d}{dn}s_n=-\frac{1}{n^2}.$$

Therefor the original sum diverges. Is this proof correct? Is there a simpler way of showing this?

Answer 1

For small $x$, $\arctan x\approx x$ and the series diverges like that of $n^{-1/2}$. (It shouldn't be difficult to find $\alpha>0$ such that $\arctan x>\alpha x$ for $x\le 1$.

Answer 2

Since $\lim_{x\rightarrow 0^{+}}\dfrac{\tan^{-1}x}{x}=1$, so for some $c>0$, for large $n$, $\tan^{1}\left(\dfrac{1}{n^{1/2}}\right)\geq\dfrac{c}{n^{1/2}}$, so one uses comparison rule to conclude.