# Puzzles/Physics puzzles/Eleatic School/Achilles comment

One traditional answer to this problem is that in fact we only looking a constant amount of time ahead. To see this let us assume that the tortoise advances at unit speed and Achilles at speed x, starting y units behind the tortoise (say, at point 0). Then it takes Achilles ${\frac {y}{x}}$ units in time (say seconds) to reach the starting point of the tortoise. In the meantime the tortoise has advanced to the point $y+{\frac {y}{x}}$ being now ${\frac {y}{x}}$ units adhead. Since x > 1 the distance has decreased. To reach the second point of the tortoise it takes Achilles the time of ${\frac {y/x}{x}}={\frac {y}{x^{2}}}$ . By induction we have, that it takes Achilles the time of $y\sum _{i=1}^{n}{x^{-i}}$ to reach the point the tortoise reached after her $n$ th advance.
For $n\to \infty$ this expression converges to ${\frac {y}{1-{\frac {1}{x}}}}=:T$ which is clearly finite. Therefore in whole our race we are not considering points in time greater or equal than $T$ . In fact at $T$ Achilles has reached the tortoise and is going to overtake it at any infinitesimal time thereafter.