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.

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