If Achilles, without his tortoise, covers $\frac1{2^n}$ meters in $\frac1{5^n}$ seconds for each $n$, does he reach $2$ meters in $\frac54$ seconds? 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 $d=2$?

 A: Since you have assumed that you cannot jump in space, $D$ would be a continuous function of $t$ and so you can show that the limit of $D$ is $2$ as $t \to \frac54$ from below.  The continuity would imply $D=2$ when $t= \frac54$.
This is physically impractical, since there is no upper bound on the velocity as $t \to \frac54$.
Incidentally, you do not need to assume that velocity is constant during each of the time intervals.  Since at the end of each interval you have $D=2\left(1-\frac12^n\right)$ when $t=\frac54\left(1-\frac15^n\right)$ for positive integer $n$, this implies $$D=2-(5-4t)^{\log(2) / \log(5)}$$ for these $t$, and if you were to assume this was also true for all positive $n$, i.e. for all $0 \lt t \lt \frac54$, then not only would you get the same limit result since $5-4t \to 0$ but you would also get a continuous velocity over time in that interval of $v=4\frac{\log(2)}{\log(5)} (5-4t)^{\log(2) / \log(5) -1}$.  Note that this velocity also tends to $\infty$ as $t \to \frac54$ from below.   
A: Yes, with your data for $t \to 5/4$  the distance covered will be $d \to 2$ , apart from considerations on the physical realizability.
Yes, the speed will tend to  infinite , but $\Delta t \to 0$ and faster, so much so that 
$v \Delta t = \Delta d \to 0$ and $\sum v\Delta t$ remains finite and $\to 2$.
