Let $p$ be prime. The sequence
$$P_k = 1 + p + p^2 + p^3 + \cdots + p^k$$
goes to $\infty$ for $k\rightarrow\infty$ in the Euclidean norm, but in the p-adic norm it goes to $\frac{1}{1-p}$. It does so by enigmatically going through the "point at infinity" coming from the positive numbers and returning to $\frac{1}{1-p}$ through the negative numbers, which may remind one somehow of the real projective line:
[Note, that the larger $p$ the closer $\frac{1}{1-p}$ to $0$ in the Euclidean norm, but for finite $p$ the number $0$ will not be reached again.]
For the $p$th root of unity $\omega = e^{i2\pi/p}$ we have $1 + \omega + \omega^2 + \omega^3 + \cdots + \omega^{p-1} = 0$, so one might be inclined to say that the sequence
$$\Omega_k = 1 + \omega + \omega^2 + \omega^3 + \cdots + \omega^{k}$$
approaches a limit cycle for $k\rightarrow\infty$, going through $0$ again and again (the smaller $p$ the more often). So with tongue in cheek one might say, that $\Omega_k$ goes to $0$ for $k\rightarrow\infty$ in the Euclidean norm.
Which deeper connection between p-adic and complex numbers does this reveal (if any)?
[Note, that the closeness of $\frac{1}{1-p}$ to $0$ corresponds to the frequency with which $0$ is passed by $\Omega_k$ — with $\Omega_k$ depending on $p$ via $\omega = e^{i2\pi/p}$.]
[Note further, that the p-adic numbers $\mathbb{Q}_p$ are an extension of $\mathbb{Q}$ (next to $\mathbb{R}$) while the complex numbers $\mathbb{C}$ are an extension of $\mathbb{R}$ (and thus of $\mathbb{Q}$).]
[Note further, that $\{\Omega_k\} \cap \mathbb{Q}_p = \{\Omega_k\} \cap \mathbb{Q} = \{0,1\}$ for $p >2$. Or does this statement not make sense?]
Just now I've learned, that Alexander Bogomolny, creator and maintainer of the marvelous web site cut-the-knot, has passed away on the 7th of July. I am very sad about that, I did admire and learn a lot from him.

This is nonsense. In the $p$-adic numbers there is no "order" and there are no "positive numbers" and no "negative numbers".