Apparent paradox for the bird traveling between two trains puzzle Trying the "hard solution" for the puzzle below (which has been
discussed, with a different angle, elsewhere on this forum)
I got to a point where I have three seemingly valid solutions, and two do not match. I have checked the procedure several times, both by hand and with Maxima, and it appears correct - but I am obviously missing something.I cannot sort this out, maybe someone here can. (Sorry for the ASCII math, I am not MathJax-enabled)
The story and the puzzle:
A colleague approached one day John Von Neumann with a puzzle that had
two paths to a solution, a laborious, complicated calculation and an
elegant, Aha!-type solution. This colleague had a theory: in such a
case, mathematicians work out the laborious solution while the
(lazier, but smarter) physicists pause and find the quick-and-easy
solution. Which solution would von Neumann find? You know the sort of
puzzle: Two trains, 100 miles apart, are approaching each other on the
same track, one going 30 miles per hour, the other going 20 miles per
hour. A bird flying 120 miles per hour starts at train A (when they
are 100 miles apart), flies to train B, turns around and flies back to
the approaching train A, and so forth, until the two trains
collide. How far has the bird flown when the collision occurs? “Two
hundred and forty miles,” von Neumann answered almost
instantly. “Darn,” replied his colleague, “I predicted you’d do it the
hard way, summing the infinite series.” “Ay!” von Neumann cried in
embarrassment, smiting his forehead. “There’s an easy way!”
Solution 1  - Easy:
The trains will collide in $T_t=\frac{D_0}{V_1+V_2}$. Hence $D_b=V_b\cdot\frac{D_0}{V_1+V_2}=120\cdot\frac{100}{50}=240$
Hard Solution preliminary:
On the first leg, the bird travels the distance $D_0$ in time $T_0=\frac{D_0}{V_b+V_2}$;
On the second leg, distance and duration amount to:
$D_1 = D_0-T_0 \cdot (V_1+V_2) = D_0 \cdot \frac{V_b-V_1}{V_b+V_2}; T_1=D_0 \cdot \frac{V_1-V_b}{V_1+V_b}*(V_2+V_b))=T_0 \cdot \frac{V_b-V_1}{V_b+V_1}$
On the third leg,  a recurrence emerges:
$D_2 = D_0 \cdot \frac{(V_1-V_b)\cdot(V_2-V_b)}{(V_1+V_b)\cdot(V_2+V_b)} ; T_2 = T_0 \cdot \frac{(V_1-V_b)\cdot(V_2-V_b)}{(V_1+V_b)\cdot(V_2+V_b)}$
The same recurrence is found to hold between $D_3$ and $D_1$ and $T_3$ and $T_1$.
If we set:
$$R=\frac{(V_1-V_b)\cdot(V_2-V_b)}{(V_1+V_b)\cdot(V_2+V_b)}$$
We see that both the total duration and the total distance traveled by the bird can be computed summing a
geometrical series  with coefficient $R$ and respective first terms $(T_0+T_1)$ and $(D_0+D_1)$. $R< 1$ holds, so
convergence is assured.
Therefore we have.
Solution 2:
$$T_t= \sum (T_0+T_1)R^n = \frac{D_0}{V_1+V_2}; D_b= V_b\cdot\frac{D_0}{V_1+V_2}$$
This matches solution (1) as it should.
Solution 3:
$$D_b= \sum(D_0+D_1)R^n ) = D_0 \cdot \frac{(V_1+V_b)\cdot(V_2-V_1+2 V_b)}{2 V_b\cdot(V_2+V_1)}$$
I am stymied.
 A: And of course there is no paradox. (As mentioned by Ron Gordon above, The same problem/same angle is discussed on: math.stackexchange.com/questions/346384/)
Solution 3 - above - is no solution at all. It just sums the distances of the trains at the start of each leg, but that is not the distance flown by the bird on that leg: that is given by (obviously) the bird's velocity times the leg duration. Whoops. The fixed solution is below.
Solution 1  - Easy:
The trains will collide in $T_t=\frac{D_0}{V_1+V_2}$. Hence $D_b=V_b\cdot\frac{D_0}{V_1+V_2}=120\cdot\frac{100}{50}=240$
**Hard Solution **
On the first leg, the ** trains are separated by** the distance $D_0$, which the birds covers in time $T_0=\frac{D_0}{V_b+V_2}$. 
The distance flown by the bird is $S_0=V_B\cdot\frac{D_0}{V_b+V_2}$
On the second leg, distance between trains, duration and distance covered by the bird amount to:
$D_1 = D_0-T_0 \cdot (V_1+V_2) = D_0 \cdot \frac{V_b-V_1}{V_b+V_2};$
$T_1=D_0 \cdot \frac{V_1-V_b}{V_1+V_b}*(V_2+V_b))=T_0 \cdot \frac{V_b-V_1}{V_b+V_1};$ 
$S_1=V_b \cdot T_1$
On the third leg,  a recurrence emerges:
$D_2 = D_0 \cdot \frac{(V_1-V_b)\cdot(V_2-V_b)}{(V_1+V_b)\cdot(V_2+V_b)} ;$ 
$T_2 = T_0 \cdot \frac{(V_1-V_b)\cdot(V_2-V_b)}{(V_1+V_b)\cdot(V_2+V_b)}; $
$S_2=V_b \cdot T_1$
The same recurrence is found to hold between $D_3$ and $D_1$, $T_3$ and $T_1$, $S_3$ and $S_1$. 
If we set:
$$R=\frac{(V_1-V_b)\cdot(V_2-V_b)}{(V_1+V_b)\cdot(V_2+V_b)}$$
We see that both the total duration and the total distance traveled by the bird can be computed summing a 
geometrical series  with coefficient $R$ and respective first terms $(T_0+T_1)$ and $(S_0+S_1)$. $R< 1$ holds, so 
convergence is assured.
Therefore we have:
Solution 2:
$$T_t= \sum (T_0+T_1)R^n = \frac{T_0+T_1}{1-R} = \frac{D_0}{V_1+V_2}; $$ 
$$S_b= \sum (S_0+S_1)R^n = V_b\cdot T_t = V_b\cdot\frac{D_0}{V_1+V_2}$$
This matches solution (1) as it should.
