geometry - Diagonal ladder length

Let $ABCD$ be a square with side length $a$. Let $s$ be a staircase from $A$ to $C$ with total length $l$ and number of steps $n$. It consists of perpendicularly alternating lines of length $\frac{a}{n}$, as pictured here.

We see that $l$ can be expressed as follows:
$$l = \frac{a}{n} \cdot n + \frac{a}{n} \cdot n = \frac{a}{n} \cdot n \cdot 2 = 2a$$

and as such stays constant at $2a$.

Now let us imagine that the amount of steps is infinite, e.g. $n = \infty$. Per definitionem, the staircase should now be the diagonal of the square with length $l=a\sqrt{2}$ according to Pythagoras. This is paradoxical! According to the equation pictured above, it should have the length $2a$, not $a\sqrt{2}$.

My question is:

Does $l$ equal to $2a$ or $a\sqrt{2}$?