Two trains move towards each other, a bird moves between them. How many trips can the bird make? The problem statement:

Two trains move towards each other at a speed of $34\ km/h$ in the same rectilinear road. A certain bird can fly at a speed of $58\ km/h$ and starts flying from the front of one of the trains to the other, when they're $102\ km$ apart. When the bird reaches the front of the other train, it starts flying back to the first train, and so on. 



*

*How many of these trips can the bird make before the two trains meet? 

*What is the total distance the bird travels?


Commentary:
The second question of the problem seems relatively simple, since one only has to notice that the trains will take 1.5 hours to meet, therefore, the bird travels $58\cdot1.5=87 km$. However, the first question baffles me. How can one calculate how many trips the bird makes? If I'm correct, in order to obtain the time the bird will take to make its first trip, we have to add the bird's speed and the speed at which the distance of the trains is being reduced ($68\ km/h$).
This means the bird will take $\frac{102}{126}\approx0.809$ hours to finish the first trip, and the trains will be $\frac{986}{21}\approx 46.95\ km$ apart. If I continue this way (now finding how long will the bird take to travel those 46.95 km), it seems that I'll never stop or that at least it will take a huge amount of trips that cannot be computed by hand. Is there a way to find a 'quick' answer to this problem? Am I making it more complicated than it actually is? 
Thanks in advance!

In the stationary reference frame of the train the bird does not start on, they are 102 km apart and the far train moves at 68 km/h.  The bird flies at 92 km/h.
It reaches the 'stationary' train in $\frac{102}{92} h$, after which the far train has moved $68 * \frac{102}{92} km$.
If the trains started X apart, then it reaches the 'stationary' train in $\frac{X}{92} h$  and the trains are now $X-\frac{68X}{92} km = (1-\frac{68}{92}) X km$ apart.
This then applies recursively, so after $K$ flights, the trains are $102 * (1-\frac{68}{92})^K km$ apart.  For no number of flights $K$ does this reach 0, so the bird must fly an infinite number of times for the trains to collide, assuming an infinitely small bird.
For a sanity check, we'll now take this solution and see if we can get the proper flight distance from it.
On flight K, the bird has flown the distance between the cars after flight $K$, times $\frac{58}{58+34}$ to account for frame-shift.  So the bird flies the sum, from $1$ to $\infty$, of $\frac{58}{92} * 102 * (1-68/92)^K$ km.
Using the identity $1+x+x^2+x^3+... = \frac{1}{1-x}$ this is $102 km * \frac{1}{1 - (1-\frac{68}{92})} * \frac{58}{92}$, aka $102km * \frac{92}{68} * \frac{58}{92}$, or $87 km$.
