geometric progression modified

During a night out, a friend brought up an interesting topic: if you have a fixed length and you always walk the half which is left, you will never get to the end, but it's trending towards it. In fact it is written like this: $\sum\limits_{i=1}^{\infty} \frac{1}{2^i} = 1$ We can change that to a wider approach: $\sum\limits_{i=1}^{\infty} k^{-i}$. Where will you end up, if you walk: $\frac{1}{k}$ If you simulate that behaviour with a small python script you get different outcomes.

sum([ 1/k**i for i in range(1,1000)])

For $k=3 \rightarrow \sum = \frac{1}{2}$, for $k=4 \rightarrow \sum = \frac{1}{3}$. Is it save to say: For $k \rightarrow \sum = \frac{1}{k-1}$? No it's not...but i can prove that this is the outcome :) Bare with me during this prove, it will be short and painless. I think a lot of people have heard about the basic prove by indication of the geometric progression. If not...i shall repeat it! $\sum\limits_{i=0}^{N} x^i = \frac{1-x^{N+1}}{1-x}$. For $N=0$ it should be trivial. But $N \rightarrow N+1$: $\sum\limits_{i=0}^{N+1} x^i = \frac{1-x^{N+2}}{1-x}$. It is save to say $\sum\limits_{i=0}^N x^i + x^{N+1} = \frac{1-x^{N+1}}{1-x} + x^{N+1}$ because $\sum\limits_{i=0}^N x^i$ is already proofed correct :) We get $\frac{1-x^{N+1}-(1-x)x^{N+1}}{1-x}=\frac{1-x^{N+1}+x^{N+1}-x^{N+2}}{1-x}$ which leaves us with: $\frac{1-x^{N+2}}{1-x}$ q.e.d. We proofed the basic concept of this progression...but the main part is to show, that $|x| = \frac{1}{k}$ is just a special case. To proof it, we don't want to go to $N$...no we want it all the way to infinity, so: $\lim\limits_{N \rightarrow \infty} \frac{1-x^{N+1}}{1-x} = \frac{1}{1-x}$. We have to look at $\frac{1}{1-x}-1$ now, because we startet with $\sum\limits_{i=0}^\infty$. So we end up with $\frac{1}{1-x}-1 = \frac{x}{1-x}$ If we think about, we wanted $|x| < 1$, let's set $|x| = \frac{1}{k}$, we get: $\frac{\frac{1}{k}}{1-\frac{1}{k}} = \frac{1}{k-1}$ which is exactly what we wanted to proof. we see: $\sum\limits_{i=1}^{\infty} \frac{1}{k^i} = \frac{1}{k-1}$ q.e.d. What does this mean in real life: If you take a fixed length and measure everytime the third of the remaining length, you will end up at exactly half of the length. Or: if we walk $\frac{1}{k}$ of the remaining length we will end up at a fraction of $\frac{1}{k-1}$ of the original length. incredible isn't it? so long