The story was that John von Neumann was asked about the following problem :

- 2 bikes are headed toward each other at the same speed 60 km/h, they are separated each other by 2 kilometers. A Fly leaves the front of the first bike and travels at 90 km/h (a very fast one) towards the second one then back to the first bike and so on
- how far does the fly travel ?

i am going to explain the methodical way used by John von Neumann :

Bikes are traveling at the speed **1 km/minute** and the fly **1.5 km/minute**

let look at the fist trip when the fly collide with the second bike before going back to the first bike:

we call **d(Fly)**, the distance corresponding to the travel of the fly, and **d(Bike 2)** the one corresponding to the Bike 2 :

we have **d(Fly) + d(Bike 2) = d(Total) = 2 **

at time t, they collide : **t *** ** ( V(Fly) + V(Bike 2)** **)** = **d(Total)**

then **t** = **d(Total) / (**** V(Fly) + V(Bike 2)** ) = d(Total) / ( 1 + 1.5 ) = 2/5 * d(Total)

then it comes **d(Fly) = (1.5 *** **d(Total)) * 2/5 = 6/5 (1)**

and **d(Bike 2) = ****d(Total)) * 2/5 = 4/5 = d(Bike 1) (2)**

Now for the second trip the distance to go back to the first bike leaving the front of second bike is **d(Fly) – d(Bike 1) = 2/5**

but we don’t have to redo the calculation again, we have to infer 2/5 in **d(Total) **for equation (1) and (2) which is 1/5 of total distance ( 2 km ), then **d(Fly) = 6/ 5*5 **and so on :

then we can extract an infinite serie for the fly travel distance :

d(Fly n steps) = 6/5 ( 1 + 1/5 + 1/(5*5 ) + 1/ (5*5*5) + … + 1/(5*5*5 * n times))

d(Fly) = d(Fly n steps) when n → ∞

**d(Fly) = 6/5 * 1/(1 – 1/5) = 3/2 km**