Analogy between p-adic and complex numbers 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.

 A: 
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

This is nonsense.  In the $p$-adic numbers there is no "order" and there are no "positive numbers" and no "negative numbers".
A: I strongly agree with the point raised in the other answer.
The not-so-deep connection here is that we have two algebraic structures which 1) are rings which 2) contain an element $x$ such that $1-x$ is invertible, and powers of $x$ exhibit an easily understood structure. To elaborate,

*

*In any ring for any $x$, we have $(1-x)\cdot (1+x+x^2+...+x^{k}) = 1-x^{k+1}$.


*Which, if we can divide by $1-x$, means that your sums are of the general form
$$X_k = \dfrac{1-x^{k+1}}{1-x} $$
Now in one case ($x=p$), we have an extra structure on our ring w.r.t. which $p^n \to 0$, so of course $X_k$ goes to $$\dfrac{1-0}{1-x} = \dfrac{1}{1-x}.$$ In the other case $x=\omega$, periodically $x^n=1$, so with the same period, $X_k$ will be $$\dfrac{1-1}{1-x} = 0.$$
Similarly I can build more examples:

*

*In the ring $\mathbb F_{17} [Z, Y]$ (polynomials in two variables over the field with 17 elements), there exists a primitive $16$-th root of unity, call it $\zeta$. Using it as $x$, we get that $$Z_k = 1+\zeta +... + \zeta^k$$ "cycles through 0 periodically". If we now extend the base field to $\mathbb F_{17^2}, \mathbb F_{17^3}$, and take a $288$-th, ($17^3-1$)-th ... root of unity as $\zeta$, this changes the frequency of the cycle.


*On the field $\mathbb R ((t))$ (Laurent series over $\mathbb R$), there is a natural metric w.r.t. which $t^n \to 0$. So for $$T_k = 1 +t+...+t^k,$$ we have that $T_k$ goes to $\dfrac{1}{1-t}$. Would you say "this enigmatically goes through infinity and comes back from the left hand side"? If we now went to an algebraic closure of this and replaced $t$ by $t^{1/2}, t^{1/3}$, would you say "it comes closer to 0 again"?
Would you say that this shows a deeper connection between these two rings I just made up? Would you say that the powers of $17$ in the first and the roots of $t$ in the second might have a profound connection? (Sorry for teasing.)

Side note 1: Doesn't the "frequency = closeness" correspondence go exactly the wrong way? Namely, for (Euclideanly) big $p$, $1/(1-p)$ comes closer to $0$, whereas $\Omega_k$ hits $0$ less frequently.
Side note 2: To your final small remark: If you define your roots of unity as complex numbers, $\omega = e^{2\pi i /p}$, then of course they are not contained in $\Bbb Q_p$ for any $p \ge 3$. However, the field $\Bbb Q_p$ in general does contain non-trivial roots of unity, namely exactly the $(p-1)$-th ones. With those playing the role of your $\omega$ or my $\zeta$ above, you get the exact same kind of behaviour. My point is, that's because they are roots of unity, and that alone does not tell you something deep about those fields, or let's say: Nothing deeper than that these fields contain certain higher roots of unity.
