Intuition for divergence of $\arctan \left(\frac{1}{\sqrt n}\right)$, the angle in the Spiral of Theodorus A (UK sixth form; final year of high school) student of mine raised the interesting question of how to prove that the total angle in the Spiral of Theodorus (formed by constructing successive right-angled triangles with hypotenuses of $\sqrt{n}$), diverges.
He identified that this is equivalent to proving the divergence of the series $$\sum_{r=1}^\infty \arctan \left(\frac{1}{\sqrt r}\right)$$ and came up with an interesting proof attempt which didn't conceptually work (although it was very nicely thought of).
The best I could offer by way of intuition is that $\arctan\left(\frac{1}{\sqrt n}\right) \approx \frac{1}{\sqrt n}$, and the latter series diverges by comparison with $\frac{1}{n}$. But the $\arctan$ value is strictly lesser, so that doesn't convert into a precise proof as far as I can see.
He hasn't been taught formal convergence tests (and it's a while since I was taught them!) although I'm sure he'd be very open to learning. However, I can't shake the feeling there ought to be a nice geometrical demonstration that the spiral does indeed keep winding around the starting point.
 A: Rather than concentrating on the angle, consider the arc length of the spiral.
At the $n$th step, we add a right triangle of base $\sqrt n$ and hypotenuse $\sqrt{n+1}$, with the outer side (of course) of length 1 perpendicular to the base. So we add a new arc to the spiral which must be greater than 1. Hence the arc length at $n$ steps must be greater than $n$. Thus the arc length diverges, and so must the angle.
(FWIW, a decade or so ago, I spent some time trying to come up with a good approximation of $n$ given a total angle of $\theta$. It gets rather messy...).
Wikipedia has some info, including an analytic continuation of the spiral published by Davis in 2001. It also has this nice diagram.

And here's a link to their SVG version. (I'm sure it'd be easy to make a much smaller SVG).
A: It is not "intuition". By MacLaurin expansion at $x=0$ we have
$$\arctan\sqrt{x}= \sqrt{x}+O\left(x^{3/2}\right)$$
Therefore
$$\arctan\sqrt{\frac{1}{n}}\sim \sqrt{\frac{1}{n}};\quad n\to\infty$$
So the given series diverges.
edit
This is a rigorous proof of the divergence
