I am Achilles II and, on a straight line, I start running really fast:
The first $1$ meter I cover in $1$ second.
The next $\frac{1}{2}$ meters, in $\frac{1}{5}$ seconds.
The next $\frac{1}{2^2}$ meters, in $\frac{1}{5^2}$ seconds.
And so on. That is, for $n=1, 2, 3, \ldots$, I cover the each successive $\frac{1}{2^n}$ meters in $\frac{1}{5^n}$ seconds.
So, as the total time I run
$$t = 1 + \frac{1}{5} + \frac{1}{5^2} + \frac{1}{5^3} + \cdots$$
tends to $\frac{5}{4}$, the total distance I cover
$$d = 1 + \frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \cdots$$
tends to $2$.
Now, my velocity is $v = (\frac{1}{2^n})/(\frac{1}{5^n}) = (\frac{5}{2})^n$. So, as $t$ tends to $\frac{5}{4}$, $v$ increases and tends to infinity.
I also assume that I cannot “jump” in space. I move only in straight lines.
The question is:
If I can move in this way, does it follow that at time $t=\frac{5}{4}$ I am at distance $d=2$?