Will a flea jumping on a stretching rubber band reach its end? The tail of a giant kangaroo is attached by a giant rubber band to a stake in the ground. A flea is sitting on top of the stake eyeing the kangaroo (hungrily). The kangaroo sees the flea leaps into the air and lands one mile from the stake (with its tail still attached to the stake by the rubber band). The flea does not give up the chase but leaps into the air and lands on the stretched rubber band one inch from the stake. The giant kangaroo, seeing this, again leaps into the air and lands another mile from the stake (i.e., a total of two miles from the stake). The flea is undaunted and leaps into the air again, landing on the rubber band one inch further along. Once again the giant kangaroo jumps another mile. The flea again leaps bravely into the air and lands another inch along the rubber band. If this continues indefinitely, will the flea ever catch the kangaroo? (Assume the earth is flat and continues indefinitely in all directions.)
I am very confused, I think the answer is NO, because 1 mile > 1 inch. Any help?
 A: The key is that as the rubber band stretches, the portion of it behind the flea also stretches proportionately.  For e.g. if the flea is at the midpoint when the kangaroo jumps, the flea still remains at the mid point of the rubber band after the jump.  So the proportion of the rubber band the flea has covered (say $p_n$) does not change because of the kangaroo jump, but it increases when the flea jumps.  So steadily (and given enough time), the flea always catches up to the kangaroo.
More analytically, let our $\mathrm{X}$-axis be along the rubber band, and start a clock after the first kangaroo jump - we will advance the clock by one step after the kangaroo leaps next.  Then when $n=0$, the kangaroo is at $1$ mile or $x_0 = 63,360$ inches.  After the first leap of the flea, $p_0 = \frac1{63,360}$.  Now the kangaroo jumps, but $p$ remains unchanged.  So when the flea leaps next, it covers an additional $\frac1{63,360\times 2}$ proportion, so $p_1 = p_0 + \frac1{63,360\times 2}$.  Similarly in general we have
$$p_n = p_{n-1} + \frac1{63, 360 \times n} = \frac1{63,360}\left(1+\frac12+\frac13+ \cdots + \frac1n \right)$$
The sum in the brackets is the partial sum of the Harmonic series (also called the Harmonic number $H_n$), which diverges, so eventually $H_n$ becomes larger than any given real number, in particular it crosses $63,360$ for some $n$.  Hence the flea catches up to the kangaroo.
In practice the number of steps required may be a bit high.  Here I estimate $n\approx 4.5 \times 10^{27,516}$ before $H_n \ge 63, 360$, which is surely beyond the stamina of most fleas (and kangaroos) and leaping once a milli-second, well beyond the heat death of the known universe...
